Question

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. The same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?A. The impulses are the same.B. The second collision.C. The first collision.D. Cannot be determined without knowing the mass of the cart.E. Cannot be determined without knowing the rebound speed of the first collision.

Answer 1

Answer: Impulse is greater in the first case. So, option C is the correct option.

Explanation:

Case 1: Cart is travelling at 0.3 m/s and collide with an stationary object and after collision, cart rebound in opposite direction and another object remains in static condition.

Applying the conservation of linear momentum:

$m_{1} \times u_{1} + m_{2} \times u_{2} = m_{1} \times v_{1} + m_{2} \times v_{2}$

$m_{1} \times 0.3 + m_{2} \times 0 = m_{1} \times v_{1} + m_{2} \times 0$

Hence velocity of cart will rebound with the same velocity i.e. 0.3 m/s

Impulse is defined as the change in momentum

Impulse on the cart = $m_{1} \times v_{1} - m_{1} \times u_{1}$ = $m_{1} \times ((-3) - (3)) = m_{1} \times (-6)$ Kg m/s.

Case 2: Initially cart is travelling at 0.3 m/s and after collision it comes to rest.

So, change in momentum or Impulse = $m_{1} \times (0 - 3)$ = $-3 \times m_{1}$ Kg m/s.

Impulse is greater in the first case. So, option C is the correct option.

Answer 2

Answer:

The second impulse is greater then the first impulse.

(B) is correct option.

Explanation:

Given that,

In first case,

Initial speed of cart = 0.3 m/s

Final speed of cart = -0.3 m/s

in second case,

Initial speed of cart = 0.3 m/s

Final speed of cart = 0 m/s

In first case,

We need to calculate the  impulse

Using formula of impulse

$I=\Delta p$

$I=\Delta (mv)$

$I=mv-mu$

Put the value into the formula

$I=m(-0.3-0.3)$

$I= -0.6m\ kg m/s$

In second case,

We need to calculate the new impulse

Using formula of impulse

$I'=\Delta (mv')$

$I'=mv'-mu'$

Put the value into the formula

$I'=m(0-0.3)$

$I'= -0.3m\ kg m/s$

So, I'> I

Hence, The second impulse is greater then the first impulse.