1. (a) What is your mass?
   (b) What is the mass of a liter bottle of water (note: the density of water is 1 g/ml)?
   (c) Which object, you or the bottle of water, will accelerate more when given a push?
2. A ball at rest in a wagon that is also at rest. When the wagon is pulled suddenly, the ball moves toward the back of the wagon even though there is apparently no force exerted on the ball to push it backwards. Does this counter Newton’s 1^st law? Why or why not?
3. An object is at rest. Suppose the following forces are exerted on it.
   F₁ = 120 N at 0°
   F₂ = 200 N at 127°
   F₃ = 160 N at 270°
   Does the object remain at rest? Why or why not?
4. A 2-kg object is at rest. Suppose the following forces are exerted on it.
   F₁ = 120 N at 0°
   F₂ = 200 N at 127°
   F₃ = 100 N at 270°
   What is the object’s acceleration (magnitude and direction)?
5. A 2-kg ball falls off a horizontal table with an initial horizontal speed of 3 m/s. While it is falling, there is a downward force exerted on the ball (gravity).
   (a) Ignoring any other forces, in which direction will the ball accelerate?
   (b) Suppose it takes 0.6 seconds for the ball to hit the ground. What is the horizontal component of the ball’s velocity the instant before it hits the ground?
6. A 2-kg object has a velocity of 2.34 m/s toward 34°. Suppose the following forces are exerted on it.
   F₁ = 100 N at 0°
   F₂ = 200 N at 127°
   F₃ = 100 N at 270°
   What is the object’s acceleration (magnitude and direction)?
7. A 2-kg ball falls off a horizontal table with an initial horizontal speed of 3 m/s. While it is falling, there is a downward force exerted on the ball (gravity). What is the magnitude and direction of the gravitational force exerted on the ball?
8. Using the equation F_g = Gm₁m₂/r² and the values given on the inside front cover of the textbook, determine the following.
   (a) The magnitude of the gravitational force exerted on the 2-kg ball in the previous question.
   (b) The magnitude of the gravitational force exerted on the 2-kg ball if the ball was moved to a height of 6,370 km above the earth’s surface.
   (c) The magnitude of the gravitational force exerted on the moon by the earth.
   (d) The magnitude of the gravitational force exerted on the earth by the moon.
9. A massless box is on a level surface. If a force of 100 N is exerted on the box in a direction 60° below the horizontal, what is the magnitude and direction of the normal force exerted by the level surface on the box?
10. (a) Suppose the coefficient of static friction between the box and surface in the previous question is 0.3. Will the box slide across the surface? Why or why not? If not, provide the magnitude and direction of the static friction force.
    (b) What if the coefficient of static friction was 0.8? Why or why not? If not, provide the magnitude and direction of the static friction force.
    
11. A horizontal force of 14 N is applied to a 10-kg box initially at rest on a plane that is inclined at an angle of 22° above horizontal.
    (a) Determine the acceleration (magnitude and direction) of the box assuming there was no friction between the box and the incline.
    (b) Determine the minimum coefficient of static friction between the box and incline that would prevent the box from sliding.
    *(c) A 2.00-kg book is held against a rough vertical wall. If the coefficient of static friction between the book and the wall is 0.300, what force perpendicular to the wall is necessary to keep the book from sliding? As a guess, compare to the force you’d need to apply to hold it up normally (which is greater?).
12. *(a) On the lab floor, the greatest acceleration I can achieve without slipping is about 4 m/s². What is the horizontal force of friction between the floor and my shoes? My mass is 70 kg. As a guess, consider how much force is needed to get me going (consider if I was on the ice and you had to give me a push).
    *(b) On the lab floor, I can slide to a stop in 1 second when I am initially going at 2 m/s. What is the horizontal force of friction between the floor and my shoes when I am sliding? As a guess, compare to the answer obtained in part (a).
    (c) A 10-kg box is sliding along a level surface. Determine the acceleration of the box if the coefficient of kinetic friction between the box and surface is 0.3.
13. A father (mass 70 kg) and his daughter (mass 25 kg) are facing each other on ice skates. With their hands, they push off against one another.
    (a) If F₁ is the force exerted on the daughter by the father and F₂ is the force exerted on the father by the daughter, what is the value of the ratio F₁/F₂?
    (b) If a₁ is the resulting acceleration of the daughter and a₂ is the resulting acceleration of the father, what is the value of the ratio a₁/a₂?
14. (a) A 0.3-kg tennis ball is hit with a 1-kg tennis racket. At the moment of contact, the tennis racket is moving at a speed of 25 m/s and slowing down. If F₁ is the force exerted on the ball by the racket and F₂ is the force exerted on the racket by the ball, what is the value of the ratio F₁/F₂?
    (b) At this very moment, the earth is orbiting the sun at a speed of about 30 km/s. If F₁ is the force exerted on the earth by the sun and F₂ is the force exerted on the sun by the earth, what is the value of the ratio F₁/F₂?
15. A 2000-kg truck is traveling eastward at 30 m/s and accelerating at a rate of 2 m/s². The truck is pulling a 500-kg trailer via a strong massless string.
    (a) If F₁ is the force exerted on the string by the trailer and F₂ is the force exerted on the string by the truck, what is the value of the ratio F₁/F₂?
    (b) If F₁ is the tension force exerted on the trailer by the string and F₂ is the tension force exerted on the truck by the string, what is the value of the ratio F₁/F₂?
16. (a) A cable supports an elevator that weighs 8000 N. What is the tension T in the cable when the elevator accelerates upward at 1.50 m/s²?
    *(b) A pendulum is placed in a car at rest and hangs vertically. The car then accelerates forward and the pendulum bob is observed to move backward, the string making an angle of 15.0° with the vertical. Find the acceleration of the car.
    (c) Two block m₁ = 20.0 kg and m₂ = 10.0 kg are connected as shown on a frictionless plane. The angle q = 25.0° and f = 35.0°. Find the acceleration of each block and the tension in the connecting string.

17. Billiard ball 2 (see figure) is at rest when it is hit with a glancing collision by ball 1 moving at a velocity of 50.0 cm/s toward the right. After the collision ball 1 moves off at an angle of 35.0° from the original direction while ball 2 moves at an angle of 40.0°, as shown in the diagram. The mass of each billiard ball is 0.017 kg.
    (a) What is the initial momentum of ball 1 (magnitude and direction)?
    (b) What is the initial momentum of ball 2 (magnitude and direction)?
18. If positive is toward the right and toward the top of the page, determine the sign of each of the following from problem #17. Explain your reasoning.
    (a) The change in ball 2’s velocity toward the right (i.e., the component of the velocity toward the right).
    (b) The change in ball 1’s velocity toward the right (i.e., the component of the velocity toward the right).
    (c) The change in ball 2’s velocity toward the top of the page (i.e., the component of the velocity toward the top of the page).
    (d) The change in ball 1’s velocity toward the top of the page (i.e., the component of the velocity toward the top of the page).
19. (a) What is the total momentum (magnitude and direction) of both balls in problem #17 before the collision?
    (b) What is the total momentum (magnitude and direction) of both balls in problem #17 after the collision?
20. *(a) Determine the speed of each ball in problem #17 after the collision.
    *(b) An 80.0-kg astronaut pushes herself away from a 1200-kg space capsule at a velocity of 3.00 m/s. Find the recoil velocity of the space shuttle.
    *(c) Two 1000-kg cars, each traveling at 20 m/s, are coming directly toward each other. If the cars stick together after the collision, how fast do they travel after the collision? (assume no external forces acting)
    *(d) A 10-kg object is moving at 10 m/s toward the right on a frictionless surface. It then collides with a 5-kg object that was at rest. After the collision, the 5-kg object moves toward the right with a speed that is 3 m/s faster than the 10-kg object’s speed toward the right. What is the speed of the 5-kg object after the collision?
    

1. (a) My mass is 70 kg. You should be able to get a good estimate of your own mass using the conversion given on the inside front cover of the textbook (i.e., 1 kg has a weight of 2.205 lb where the acceleration of gravity is 9.8 m/s²).
   (b) One liter contains 1000 ml. Thus, with a density of 1 g/ml, the mass of 1 liter is 1000 g (or 1 kg).
   (c) Since the bottle of water has less mass, it will accelerate more with the same force exerted upon it.
2. No, it doesn’t counter Newton’s 1^st law because the motion toward the back is relative to the motion of the wagon, which is accelerating and thus does not provide an initial reference frame. Relative to the floor, which is not accelerating, the ball appears to try to stay where it was, which is consistent with Newton’s 1^st law.
3. Separate each force into two components. Here, I’ll separate into 0° and 90° components.
   F₁ = 120 N at 0° and 0 N at 90°.
   F₂ = (200 N)´ cos(127°) at 0° and (200 N)´ sin(127°) at 90°
   F₃ = (160 N)´ cos(270°) at 0° and (160 N)´ sin(270°) at 90°
   Adding up components gives:
   S F_0° = 120 N + (200 N)´ cos(127°) + (160 N)´ cos(270°) = -0.363 N
   S F_90° = 0 N + (200 N)´ sin(127°) + (160 N)´ sin(270°) = -0.2729 N
   Rounding (according to significant digits) gives components of zero. Thus, the object remains at rest (i.e., no net force).
4. Separate each force into two components. Here, I’ll separate into 0° and 90° components.
   F₁ = 120 N at 0° and 0 N at 90°.
   F₂ = (200 N)´ cos(127°) at 0° and (200 N)´ sin(127°) at 90°
   F₃ = (100 N)´ cos(270°) at 0° and (100 N)´ sin(270°) at 90°
   Adding up components gives:
   S F_0° = 120 N + (200 N)´ cos(127°) + (100 N)´ cos(270°) = 0 N
   S F_90° = 0 N + (200 N)´ sin(127°) + (100 N)´ sin(270°) = 60 N
   Combining back together gives a net force of 60 N toward 90°. Since the net force exerted on the object equals the object’s mass times its acceleration, its acceleration must be the net force divided by its mass, i.e., (60 N)/(2 kg) = 30 m/s². The direction is in the same direction as the total force, the acceleration is also toward 90°.
5. (a) Since the only force is due to gravity and it acts downward, the ball will accelerate downward.
   (b) Since the ball will accelerate downward and not horizontally, the horizontal component of the ball’s velocity remains the same, i.e., 3 m/s.
6. The forces are the same as in problem #4 except that F₁ has a magnitude of 100 N instead of 120 N. Thus, the component toward 0° will be 20 N less than before. Since it was 0 N before, it must now be –20 N, i.e.,
   S F_0° = -20 N
   S F_90° = 60 N.
   Combining back together gives a net force of 63 N toward 108° (the inverse tangent gives 18° or 72°, depending on how you do it, which corresponds to 108° in this problem). Its acceleration must be the net force divided by its mass, i.e., (63 N)/(2 kg) = 32 m/s². The direction is in the same direction as the total force. The acceleration is also toward 108°. Note that the velocity of the object is irrelevant.
7. A falling object accelerates at 9.8 m/s². From Newton’s 2^nd law, the net force exerted on it must be equal to its mass (2 kg) times its acceleration (9.8 m/s²). Thus, the net force must be 19.6 kg× m/s² or 19.6 N. The direction of the force is the same as the acceleration, i.e. downward.
8. (a) F_g = (6.67´ 10^-11 N× m²/kg²)´ (2 kg)´ (5.98´ 10²⁴ kg)/(6.38´ 10⁶ m)². The numbers were obtained from the inside front cover of the textbook for the universal gravitational constant, the mass of the earth and the radius of the earth. The radius of the earth was used because that is the distance between the center of the ball and the center of the earth. Calculating, I get 19.6 N (the same as in problem #7.
   (b) The same procedure as in (a) can be used except with a distance of 2´ 6.38´ 10⁶ m. Doubling the distance should result in the force decreasing by a factor of ¼ since r is in the denominator and squared. Calculate it just to make sure.
   (c) The same procedure as in (a) can be used except with a distance of 3.85´ 10⁸ m and the mass of the moon (7.35´ 10²² kg) instead of the mass of the ball. Doubling the distance is 60.4 times greater than that in (a), the force should be smaller by a factor of 1/(60.4²). However, the mass is about 3.7´ 10²² times greater, which means the force should be 3.7´ 10²² times greater. The net effect is that the force should be greater by a factor of (3.7´ 10²²)/(60.4²) = 1.0´ 10¹⁹, i.e. the force should be 19.6´ 10¹⁹ N. Calculate it just to make sure.
   (d) The magnitude of the gravitational force exerted on the earth by the moon should be the same as that calculated in (d). The masses are switched by the product of the masses are the same.
9. The 100-N force exerts both a horizontal force [(100 N)´ cos(60°)] and a vertical force [(100 N)´ sin(60°)] on the box. Since the box does not sink into the surface, the surface must be supplying an upward force equal to the downward force. Assuming the box has no mass (so that there is not an additional gravitational force), the normal force must be equal to [(100 N)´ sin(60°)].
10. (a) If the coefficient of static friction is 0.3, then the frictional force is 0.3 of the normal force or (0.3)´ (100 N)´ sin(60°) which can be as much as 26 N and opposes the motion (toward 180°). Since there is an applied horizontal force of 50 N, the box will slide.
    (b) (a) If the coefficient of static friction is 0.8, then the frictional force is 0.8 of the normal force or (0.8)´ (100 N)´ sin(60°) which can be as much as 69 N and opposes the motion. Since there is an applied horizontal force of only 50 N, the box will not slide. The frictional force in this case is 50 N (it does not need to be 69 N) and is directed at 180°.
11. A horizontal force of 14 N is applied to a 10-kg box initially at rest on a plane that is inclined at an angle of 22° above horizontal.
    (a) First draw a force diagram. There are three forces acting. A horizontal applied force (14 N), a vertical gravitational force (10 kg ´ 9.8 m/s²) and a normal force perpendicular to the plane. As for any vector problem, one should separate each force into two components. Here, I’ll separate into components along the plane and perpendicular to the plane rather than strictly horizontal and vertical. Although this means that the force of gravity now has to be split into two components, I know that there is no acceleration perpendicular to the plane and that helps me in the end.
    F_applied = (14 N)´ cos(22°) up the plane and (14 N)´ sin(22) into the plane.
    F_g = (98 N)´ sin(22°) down the plane and (98 N)´ cos(22°) into the plane.
    F_N = 0 N up the plane and F_N out of the plane.
    Adding up components gives:
    S F_{up the plane} = (14 N)´ cos(22°) - (98 N)´ sin(22°) = - 27.3 N (i.e. it is directed down the plane).
    S F_{out of the plane} = - (14 N)´ sin(22°) - (98 N)´ cos(22°) + F_N = F_N - 96.1 N.
    From Newton’s 2^nd law, the acceleration up the plane must be equal to the net force up the plane divided by the mass, i.e., (- 27.3 N)/(10 kg) = -2.73 m/s² up the plane (or +2.73 m/s² down the plane). Due to the normal force, there is no acceleration perpendicular to the plane. Thus, this is the total acceleration.
    (b) The approach is the same as before but an additional force is added to the force diagram, i.e., a frictional force F_f directed against the motion. Separating into components gives:
    F_f = F_f up the plane. and 0 N into the plane.
    Note that the frictional force is directed up the plane (since the motion would otherwise be down the plane). We want there to be no acceleration along the plane, thus the frictional force must be equal to 27.3 N. The coefficient of static friction gives the ratio of the frictional force to the normal force, thus the coefficient would have to be (27.3 N)/F_N. The normal force can be determined because there is no acceleration perpendicular to the plane, i.e. the net force out of the plane must be zero. Since the net force out of the plane is F_N – 96.1 N, the normal force must be 96.1 N. Thus, the coefficient is (27.3 N)/(96.1 N) = 0.28.
    (c) Again, first draw a force diagram. There are four forces acting. A horizontal applied force, a vertical gravitational force (2 kg ´ 9.8 m/s²), a normal force perpendicular to the wall and a frictional force opposing motion along the wall. As for any vector problem, one should separate each force into two components. Here, I’ll separate into components along the plane and perpendicular to the plane (which happens to be vertical and horizontal in this case)
    F_applied = 0 N along the wall and F_applied into the wall.
    F_g = (19.6 N) down the wall and 0 N into the wall.
    F_N = 0 N down the wall and F_N out of the wall.
    F_f = F_f up the wall and 0 N out of the wall.
    Adding up components gives:
    S F_{up the wall} = F_f – 19.6 N.
    S F_{out of the wall} = F_N - F_applied.
    From Newton’s 2^nd law, the accelerations up the wall and out of the wall must be zero if the book remains at rest, i.e.,
    S F_{up the wall} = F_f – 19.6 N = 0
    S F_{out of the wall} = F_N - F_applied = 0.
    Furthermore, we know that the frictional force must be related to the normal force as follows:
    F_f = 0.300´ F_N.
    This gives us three equations and three unknowns. From the first, F_f is found to be 19.6 N. From the last, F_N is found to be (19.6 N)/(0.300). From the middle, F_applied is found to be (19.6 N)/(0.300) = 65.3 N.
12. (a) The only horizontal force acting is the frictional force. Thus,
    F_f = ma.
    Since my mass is given (70 kg) and the acceleration is given (4 m/s²), the frictional force is (70 kg)´ (4 m/s²) = 280 N.
    (b) With the time and change in velocity given, the acceleration is known, i.e., D v/D t = (2 m/s)/(1 s) = 2 m/s². As in part (a), the force of friction is the product, i.e., (70 kg)´ (2 m/s²) = 140 N.
    (c) First draw a force diagram. There are three forces acting. A vertical gravitational force (10 kg ´ 9.8 m/s²), a normal force perpendicular to the surface and a frictional force opposing motion along the surface. As for any vector problem, one should separate each force into two components. Here, I’ll separate into components along the plane and perpendicular to the plane (which happens to be horizontal and vertical in this case)
    F_g = (98 N) into the surface and 0 N along the surface.
    F_N = F_N out of the surface and 0 N along the surface
    F_f = 0 N into the surface and F_f along the surface.
    Adding up components gives:
    S F_{out of the surface} = F_N – 98 N.
    S F_{along the surface} = F_f.
    The acceleration out of the surface is zero and the acceleration along the surface is unknown, so
    S F_{out of the surface} = F_N – 98 N = 0
    S F_{along the surface} = F_f = (10 kg)a_{along the surface}.
    The frictional force must be related to the normal force as follows:
    F_f = 0.3´ F_N.
    This gives us three equations and three unknowns. From the first, F_N is found to be 98 N. From the last, F_f is found to be (0.3)´ (98 N). From the middle, a_{along the surface} is found to be (0.3)´ (98 N)/(10 kg) = 2.9 m/s².
13. (a) By Newton’s 3^rd law, the two forces must be equal and opposite. The ratio is –1.
    (b) Since each force equals ma, the ratio of the forces must equal (m₁a₁)/(m₂a₂). Since this equals –1,
    (m₁a₁)/(m₂a₂) = –1,
    the ratio of the accelerations, a₁/a₂ = – m₂/m₁ = – (70 kg)/(25 kg) = 2.8. Note that the daughter accelerates more since her mass is less.
14. (a) The information regarding the speed and acceleration are irrelevant. The ratio of the forces is still –1 (i.e., the forces are equal and opposite).
    (b) Again, the information regarding the speed and acceleration are irrelevant. The ratio of the forces is still –1 (i.e., the force exerted on the sun by the earth is equal and opposite to the force exerted on the earth by the sun).
15. A 2000-kg truck is traveling eastward at 30 m/s and accelerating at a rate of 2 m/s². The truck is pulling a 500-kg trailer via a strong massless string.
    (a) Since the string is massless, the product of mass and acceleration must be zero and therefore, by Newton’s 2^nd law, the net force must also equal zero. Consequently, F₁ must be equal and opposite to F₂. Note that Newton’s 3^rd law is not invoked here. We are only talking about forces exerted on the string, not the counter-forces exerted by the string on everything else.
    (b) Here we invoke Newton’s 3^rd law. The forces exerted by the string must be equal and opposite to the forces exerted on the string. Thus, the tension force exerted on the trailer is equal and opposite to the tension force exerted on the truck.
16. (a) According to Newton’s 2^nd law, the net force on the elevator must equal to the mass of elevator times its acceleration. Its mass is its weight divided by g, i.e., (8000 N)/(9.8 m/s²). Its acceleration is 1.50 m/s². Thus, the net force must be [(8000 N)/(9.8 m/s²)]´ [1.50 m/s²] = 1200 N. This is the net force between gravity pulling downward and the tension pulling upward. Thus, the tension must be 1200 N more than gravity, i.e., T = 9200 N.
    (b) Draw a force diagram. There are only two forces acting – the tension in the string and gravity. Since the bob does not accelerate vertically, the gravity force down (mg) must equal to the component of the tension force up [ T´ cos(18°) ]. The horizontal tension force must equal its mass times its acceleration [ ma ]. At first glance, this might appear to be unsolvable because there are three unknowns (m, a and T). However, the mass will cancel out (i.e., the answer is independent of the mass). Solve the vertical equation for the tension and then plug that into the second equation. This should leave only one unknown, acceleration.
    (c) Draw a force diagram for each block. Separate each force into two components, one along the plane and one perpendicular to the plane. This should give you four equations:
    For block 1:
    S F_{up the plane} = T – F_1g´ sinq = m₁a S F_{out of the plane} = F_1N – F_g´ cosq = 0.
    For block 2:
    S F_{down the plane} = F_2g´ sinf – T = m₂a S F_{out of the plane} = F_2N – F_g´ cosf = 0
    Note that I used the same a for each. This is because they are moving together. Also note that for block 1 I used positive as up the plane whereas for block 2 I used positive as down the plane. This is because I am defining positive as the direction of a. Finally note that I used the same T for both tensions (since the string is assumed to be massless).
    In this problem, there are four equations and four unknowns (T, a, F_1N and F_2N). Thus, it can be solved for all of the unknowns.
17. Billiard ball 2 (see figure) is at rest when it is hit with a glancing collision by ball 1 moving at a velocity of 50.0 cm/s toward the right. After the collision ball 1 moves off at an angle of 35.0° from the original direction while ball 2 moves at an angle of 40.0°, as shown in the diagram. The mass of each billiard ball is 0.017 kg.
    (a) The initial momentum of ball 1 is its mass times its velocity, i.e., (0.0117 kg)´ (0.500 m/s) = 0.00585 kg× m/s = 5.85´ 10^-3 kg× m/s directed toward the right (same direction as velocity).
    (b) Ball two is initially at rest. Thus, its momentum is zero.
18. (a) Ball 2’s velocity increases toward the right, i.e., the change is positive toward the right.
    (b) Ball 1’s velocity decreases toward the right, i.e., the change is negative toward the right.
    (c) Ball 2’s velocity increases toward the bottom of the page, i.e., the change is negative toward the top of the page.
    (d) Ball 1’s velocity increases toward the top of the page, i.e., the change is positive toward the top of the page.
19. (a) Initially, ball 1 had a momentum of 5.85´ 10^-3 kg× m/s toward the right (see problem 17) and ball 2 had zero momentum. Thus, the total momentum was 5.85´ 10^-3 kg× m/s toward the right. (b) Since momentum is conserved in this collision, the total momentum after the collision is still 5.85´ 10^-3 kg× m/s toward the right.
20. (a) Since momentum is a vector, there are two equations:
    p_1ix + p_2ix = p_1fx + p_2fx
    p_1iy + p_2iy = p_1fy + p_2fy.
    Plugging in the numbers gives
    (0.0117 kg)´ (0.50 m/s) + 0 = (0.0117 kg)´ v_1fx + (0.0117 kg)´ v_2fx
    0 + 0 = (0.0117 kg)´ v_1fy + (0.0117 kg)´ v_2fy.
    Since the angles are given, these can be rewritten as
    (0.0117 kg)´ (0.50 m/s) + 0 = (0.0117 kg)´ v_1f´ cos(35°) + (0.0117 kg)´ v_2f´ cos(40°)
    0 + 0 = (0.0117 kg)´ v_1f´ sin(35°) + (0.0117 kg)´ v_2f´ sin(40°).
    Thus, there are two equations and two unknowns. One can then solve for the two unknowns.
    (b) The change in momentum of the astronaut is (80.0 kg)´ (3.00 m/s) = 240 kg× m/s. Thus must also be the change in momentum of the space shuttle, i.e., 240 kg× m/s = (1200 kg)´ D v. Solving for v gives a recoil velocity of 0.2 m/s.
    (c) Before the collision, the total momentum is zero. Consider one car moving toward the right and the other moving toward the left. In that case, one car has a momentum of (1000 kg)´ (20 m/s) toward the right and the other a momentum of (1000 kg)´ (20 m/s) toward the left. Momentum is a vector. Thus, they cancel and the total momentum is zero beforehand. When they collide, they stick together. The total momentum after the collision must also be zero so the velocity of the two stuck together must likewise be zero.
    (d) The total momentum before the collision is (10 kg)´ (10 m/s) toward the right. Afterwards, the momentum equals (10 kg)´ v₁ + (5 kg)´ v₂. These two must be equal, i.e.,
    (10 kg)´ (10 m/s) toward the right = (10 kg)´ v₁ + (5 kg)´ v₂.
    It is given that v₁ = v₂ – 3 m/s. This gives two equations and two unknowns.