Interactions. A 1.0-kg cart is initially moving to the right (positive x-directi

Question

Interactions. A 1.0-kg cart is initially moving to the right (positive x-direction) on a low friction track at 2.5 m/s. It hits a stationary 3.0-kg cart.

a. Assuming that it is an elastic collision, what is the final velocity of each cart?

b. What is the change in kinetic energy, Kf Ki , for the system of two carts in the elastic collision?

c. Suppose that in the collision, there is a spring between the two carts that makes it an elastic collision. The carts compress the spring and momentarily come to rest, before moving off at their respective velocities. How much energy is stored in the spring at the point where the carts are stationary?

d. If it takes 50 ms for the 1.0-kg cart to come to rest, what is the total force that it exerts on the spring?

Explanation / Answer

(A) for elastic collision,

vlocity of approach = vlocity of separation

2.5 = v1 - v2 .......... (i)

Applying momentum conservation,

1 x 2.5 = 3v1 + v2

3v1 + v2 = 2.5 .........(ii)

Solving ,

v1 = 5/4 = 1.25 m/s

v2 = -5/4 = - 1.25 m/s

3kg cart: -> 1.25 m/s in +ve x direction

1kg cart: 1.25 m/s in -ve x direction

(B) Kf = 1 x 1.25^2 /2 + 3 x 1.25^2 /2 = 3.125 J

Ki = 1 x 2.5^2 /2 = 3.125 J

Kf - Ki = 0

(C) 1 x 2.5 = (3 + 1) v

v = 0.625 m/s

energy stored in spring = 3.125 - (4 x 0.625^2 / 2)

= 2.34 J

(D) v = u + a t

0 = 2.5 + (50 x 10^-3) a

a = 50 m/s^2

F = m a = 50 N

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