Question

A boy of mass 50 kg is running with a velocity of 2m/s. He jumps over a stationary cart of 2 kg while running . Find the velocity of cart after jumping of the body.

Answer 1

M = mass of the boy = 50 kg

V = initial velocity of the boy before jumping into cart = 2 m/s

m = mass of stationary cart = 2 kg

v = velocity of stationary cart = 0 m/s

V' = velocity of boy and cart combination after collision

Using conservation of momentum

M V + m v = (M + m) V'

(50) (2) + (2) (0) = (50 + 2) V'

V' = 100/52

V' = 1.92 m/s

Answer 2

Answer:

V = 1.92 m/s

Explanation:

It is given that,

Mass of the boy, m = 50 kg

Velocity of the boy, v = 2 m/s

Mass of the cart, m' = 2 kg

Let V is the velocity of cart after jumping of the body. It can be calculated using the conservation of momentum as :

$mv+m'v'=(m+m')V$

v' = 0 (as the cart is stationary)

$mv=(m+m')V$

$V=\dfrac{mv}{(m+m')}$

$V=\dfrac{50\times 2}{(50+2)}$

V = 1.92 m/s

So, the velocity of cart after jumping of the body is 1.92 m/s. Hence, this is the required solution.