Question

A 0.5-kg cart moving on a horizontal frictionless track at 0.6 m/s hits a solid bumber and bounces back with a speed of 0.4 m/s. The impulse on the cart is?

Answer 1

The impulse on the cart is calculated using the change in momentum formula, resulting in an impulse of 0.5 kg·m/s directed opposite to the cart's initial direction.

Step-by-step explanation:

The student's question involves calculating the impulse on a cart after it bounces off a bumper on a horizontal frictionless track. Impulse is defined as the change in momentum of an object when a force is applied over a certain period of time. Here, impulse can also be calculated using the final and initial velocities of the cart since impulse is equal to the change in momentum.

To find the impulse, you must consider the momentum before and after the collision. The formula for impulse (I) is I = Δp = m(v_f - v_i), where m is the mass of the cart, v_f is the final velocity, and v_i is the initial velocity. Substituting the given values, we get I = 0.5 kg (0.4 m/s - (-0.6 m/s)) = 0.5 kg • 1.0 m/s = 0.5 kg·m/s. Thus, the impulse on the cart is 0.5 kg·m/s directed opposite to the cart's initial direction.

Answer 2

I = 0.5 kg.m/s

Step-by-step explanation:

given,

mass of the cart = 0.5 Kg

initial speed of cart, u= 0.6 m/s

final speed of cart, v = - 0.4 m/s

impulse of the cart = ?

impulse is equal to the change of the momentum of the cart.

I = m u - m v

I = m (u - v)

I = 0.5 (0.6 - (-0.4))

I = 0.5 x 1

I = 0.5 kg.m/s

impulse on the cart is equal to 0.5 kg.m/s