1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A block of mass 1 kg travelling in a straight line with a velocity of 10 msâˆ’1 collides with and sticks to a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact, just after the impact and the velocity of the combined object.

A
10 kg m s1, 20 kg m s1, 1.67 ms1
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
10 kg m s1, 10 kg m s1, 2 ms1
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
10 kg m s1, 10 kg m s1, 1.67 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5 kg m s1, 10 kg m s1, 2 ms1
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C 10 kg m s−1, 10 kg m s−1, 1.67 ms−1Given, Mass of first block, m1=1 kg Mass of second block, m2=5 kg Initial velocity of first block, u1=10 m/s Initial velocity of second block, u2=0 m/s The momentum before collision is: pi=m1u1+m2u2 pi=1×10+5×0 pi=10 kg m/s Let the final velocity of the blocks be v. Final momentum will be: pf=(m1+m2)v pf=(1+5)v=6v From the law of conservation of momentum, momentum before collision should be equal to momentum after the collision as no external force was acting on the bodies. So, pf=pi 6v=10 v=1.67 m/s

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Conservation of Momentum
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program