  Question

# A block  of mass 1 kg travelling in a straight line with a velocity of 10 ms−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 s120 kg m s11.67 ms1
B
10 kg m s110 kg m s12 ms1
C
10 kg m s110 kg m s11.67 ms1
D
5 kg m s110 kg m s12 ms1

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   