Question

A 75 kg man rides on a 39 kg cart moving at a velocity of 2.3 m/s. He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?

Answer 1

`\Delta v = 4.42 m/s`

Step-by-step explanation:

As we know that man and cart is taken as a system then the net force of them is zero.

So here we can apply momentum conservation for the system

So here we have

`(m_1 + m_2) v_i = m_1v_1 + m_2v_2`

now we know that

mass of man = 75 kg

mass of cart = 39 kg

initially man and cart is moving together with speed 2.3 m/s

final speed of man with respect to ground is zero

`(75 + 39) (2.3) = (75)0 + 39 v`

So here we have

`262.2 = 39 v`

`v = 6.72 m/s`

So change in velocity of the cart is given as

`\Delta v = v_f - v_i`

`\Delta v = 6.72 - 2.3 = 4.42 m/s`