Question

A block of mass $1kg$ travelling in a straight line with a velocity of $10m{s}^{–1}$ collides with and sticks to a stationary wooden block of mass $5kg.$ 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 ms^-1, 20 kgms^-1, 1.67 ms^-1
• B. 10 kg ms^-1, 10 kg ms^-1, 2 ms^-1
• C. 10 kg ms^-1, 10 kg ms^-1, 1.67 ms^-1
• D. 5 kg ms^-1, 10 kg ms^-1, 2 ms^-1

Answer 1

The correct option is C

10 kg ms^-1, 10 kg ms^-1, 1.67 ms^-1

Given, $m_1=1kg,u_1=10m/s$ and $m_2=5kg,u_2=0m/s$

The momentum before collision $=m_1{u}_1+m_2{u}_2=10kgm/s$

The momentum after collision $=(m1+m2)v=6v$.

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,

$6v=10$

$\Rightarrow v=1.67m/s$