Question

A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of +0.40 m/s, and cart A is initially at rest. After the collision the x component of velocity of the standard cart is +0.20 m/s and the x component of velocity of cart A is +0.65 m/s . After the collision, cart A continues to the end of the track and rebounds with its speed unchanged. Before the carts collide again, you drop a lump of putty onto cart A, where it sticks. After the second collision, the x component of velocity of the standard cart is -0.20 m/s and the x component of velocity of cart A is +0.50 m/s .

1. What is the inertia of the putty?

Answer 1

mass of standard cart = 1kg

mass of cart A

initial velocity of standard cart = +0.4 m/s

initial velocity of cart A = 0 m/s

final velocity of standard cart after 1^st collision = +0.2 m/s

final velocity of cart A after 1^st collision = +0.65 m/s

No external force is acting on the system.

So, momentum is conserved

Initial momentum = final momentum

∴

∴

Let the velocity of cart A after rebound be

After rebound, speed is same and direction is reversed

∴

Let the mass of putty be

Let initial velocity of putty =

Since the putty sticks to cart A, they will have common velocity after collision.

Let this velocity be

Again, conserving momentum,

∴

∴

Now the cart A + putty system and standard cart again collide,

final velocity of cart A + putty system after final collision = + 0.5 m/s

final velocity of standard cart after final collision = - 0.2 m/s

Again, conserving momentum, we get

∴

∴

So, mass of putty is 0.9923 kg