A boy of 50 kg is ruining with a velocity of 2m/s he jumps over a stationary of cart of 2kg while running find the velocity of cart after jumping of boy.

3.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 1.9
Formula $M_{1} u_{1} + m_{2} u_{2} = (M + m) v_{2}$
$M = 50 k g, u_{1} = 2 m / s, m = 2 k g, u_{2} = 0$
After jumping on to cart both man and cart move with common velocity $.$
So $,$ by substituting values we get $,$
$50 (2) + 2 (0) = (50 + 2) v_{2}$
$100 + 0 = 52 v_{2}$
$v_{2} = 100 / 52 = 1.9$
Hence,
option $(C)$ is correct answer.

Suggest Corrections

Similar questions

A boy of $50 k g$ mass is running with a velocity of $2 m s^{- 1}$ . He jumped into a stationary cart of $2 k g$ while running. Find the velocity of the cart after the boy jumped into it.

Q. A boy is running with a velocity of

2 m s^{- 1}

. He jumps over a stationary cart of

2 k g

while running. If the mass of the boy is of

20 k g

, find the velocity at which the cart moves after he jumps over the cart. (Neglect friction between the cart and ground)

A boy of 50 kg mass is running with a velocity of 2 m/s. He jumped into a stationary cart of 2 kg while running. Find the velocity of cart after the boy jumped into. (in m/s)

Q. A boy is running with a velocity of

2 m s^{- 1}

. He jumps over a stationary cart of

2 k g

while running. If the mass of the boy is of

20 k g

, find the velocity at which the cart moves after he jumps over the cart. (Neglect friction between the cart and ground)

A boy of $50$ kg mass is running with a velocity of $2$ $\frac{m}{s}$ . He jumped into a stationary cart of $2$ kg while running. Find the velocity of the cart after the boy jumped into it. (in $\frac{m}{s}$ )

Join BYJU'S Learning Program

A boy of 50 kg is ruining with a velocity of 2m/s he jumps over a stationary of cart of 2kg while running find the velocity of cart after jumping of boy.

The correct option is C 1.9Formula M1u1+m2u2=(M+m)v2M=50kg,u1=2m/s,m=2kg,u2=0After jumping on to cart both man and cart move with common velocity.So, by substituting values we get,50(2)+2(0)=(50+2)v2100+0=52v2v2=100/52=1.9Hence,option (C) is correct answer.