wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A boy is running with a velocity of 2 ms−1. He jumps over a stationary cart of 2 kg while running. If the mass of the boy is of 20 kg, find the velocity at which the cart moves after he jumps over the cart. (Neglect friction between the cart and ground)

A
1.81 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.92 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.44 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.3 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1.81 ms1

Given mass of the boy, m1=20 kg
Initial velocity of boy, u1=2 ms1
Given mass of the cart, m2=2kg
Initial velocity of the cart, u2=0 (Initially at rest)
Let v1 be the final velocity of the boy and v2 be the final velocity of the cart.

Since,the boy jumped over the cart, the final velocity v1 of boy will be equal to that of the cart.
Therefore v1=v2
According to law of conservation of momentum:
m1u1+m2u2=m1v1+m2v2
Sustituting the values , we get,
20×2+2×0=20×v1+2×v2
40=20×v2+2×v2
40=(20+2)×v2
4022=v2
v2=1.81 ms1
Therefore, velocity of the cart after he jumps over it is equal to 1.81 ms1.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conservation of Momentum
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon