# A 2.0 kg cart moving to the right with a speed of 10 m/s

Please see the attachment.

11. A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to the left with a speed of 2 m/s. After the collision, the 3.0-kg cart is moving to the left with a speed of 1 m/s. What is the final velocity of the 5.0-kg cart?

(a) zero m/s

(b) 0.8 m/s to the right

(c) 0.8 m/s to the left

(d) 2.0 m/s to the right

(e) 2.0 m/s to the left

12. A 1000-kg car traveling east at 20 m/s collides with a 1500-kg car traveling west at 10 m/s. The cars stick together after the collision. What is their common velocity after the collision?

(a) 16 m/s, east

(b) 6 m/s, west

(c) 4 m/s, east

(d) 2 m/s, east

(e) 1 m/s, west

13. A 50.0-kg boy runs at a speed of 10.0 m/s and jumps onto a cart as shown in the figure. The cart is initially at rest. If the speed of the car with the boy on it is 2.50 m/s, what is the mass of the cart?

(a) 150 kg

(b) 175 kg

(c) 210 kg

(d) 260 kg

(e) 300 kg

14. Care One is traveling due north and Car Two is traveling due east. After the collision shown, Car One rebounds due south. Which of the numbered arrows is the only one that can represent the final direction of Car Two?

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

15. A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes into two pieces. After the explosion, a 0.20-kg piece moves south at 4.0 m/s. What are the components of the velocity of the 0.30-kg piece?

(a) 4.0 m/s north, 0 m/s

(b) 2.7 m/s north, 3.3 m/s west

(c) 4.0 m/s north, 2.7 m/s west

(d) 0 m/s, 2.0 m/s east

(e) 4.0 m/s north, 2.0 m/s east

16. In the game of billiards, all the balls have approximately the same mass, about 0.17 kg. In the figure, the cue ball strikes another ball such that it follows the path shown. The other ball has a speed of 1.5 m/s immediately after the collision. What is the speed of the cue ball after the collision?

(a) 1.5 m/s

(b) 1.8 m/s

(c) 2.6 m/s

(d) 4.3 m/s

(e) 5.2 m/s

17. A 1500-kg satellite orbits a planet in a circular orbit of radius 6.2 x 10^6 m. What is the angular momentum of the satellite in its orbit around the planet if the satellite completes one orbit every 1.5 x 10^4 s?

(a) 3.9 x 10^6 kg x m^2/s

(b) 1.4 x 10^14 kg x m^2/s

(c) 6.2 x 10^8 kg x m^2/s

(d) 8.1 x 10^11 kg x m^2/s

(e) 2.4 x 10^13 kg x m^2/s

18. A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases her moment of inertia by 50%. What is the skater's final angular speed?

(a) 3.0 rad/s

(b) 4.5 rad/s

(c) 9.0 rad/s

(d) 12 rad/s

(e) 18 rad/s

Two skaters, each of mass 40 kg, approach each other along parallel paths that are separated by a distance of 2 m. Both skaters have a speed of 10 m/s.

The first skater carries a 2-m pole that may be considered massless. As he passes the pole, the second skater catches hold of the end. The two skaters then go around in a circle about the center of the pole.

19. What is the angular speed of the skaters after they have linked together?

(a) 4 rad/s

(b) 5 rad/s

(c) 10 rad/s

(d) 20 rad/s

(e) 40 rad/s

20. What is their combined angular momentum about the center of the pole?

(a) 2 kg x m^2/s

(b) 40 kg x m^2/s

(c) 80 kg x m^2/s

(d) 400 kg x m^2/s

(e) 800 kg x m^2/s

#### Solution Preview

Hello,

Most of the problems are dealing with conservation of momentum. The momentum of an object of mass m, moving with a velocity v is given as m x v.

Conservation of momentum in very simple terms means that :

Total momentum before collision (or explosion) = Total momentum after collision (or explosion).

The trick in solving such types of problems is to always keep in mind that momentum is a vector and so direction is VERY IMPORTANT.

11) Let the 3.0kg mass be 1 and the 5.0kg mass be 2. Also let velocity to the right be positive and that to the left be negative.

m1 = 3.0kg m2 = 5.0kg

u1 = 1.0m/s u2 = -2m/s (because it is to the left)

v1 = -1m/s v2 = ?

Writing the law of conservation of momentum mathematically we have:

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2 (u refers to the initial velocities while v refers to the final velocities)

from here we get that

v2 = (m1 x u1 + m2 x u2 - m1 x v1)/ m2

If you put in the values above you will get that v2 = -0.8m/s. Since it is negative, it means it is to the left and so the right answer is (c)

12) The same kind of problem as above. This time we take east as ...

#### Solution Summary

This in-depth solution tackles problems on the principle of conservation of momentum with step-by-step calculations and brief justifications.