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 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. $10kgm{s}^{-1}$, $20kgm{s}^{-1}$, $1.67m{s}^{-1}$
• B. $10kgm{s}^{-1}$, $10kgm{s}^{-1}$, $2m{s}^{-1}$
• C. $10kgm{s}^{-1}$, $10kgm{s}^{-1}$, $1.67m{s}^{-1}$
• D. $5kgm{s}^{-1}$, $10kgm{s}^{-1}$, $2m{s}^{-1}$

Answer 1

The correct option is C $10kgm{s}^{-1}$, $10kgm{s}^{-1}$, $1.67m{s}^{-1}$

Given,
Mass of first block, $m_1=1kg$
Mass of second block, $m_2=5kg$
Initial velocity of first block, $u_1=10m/s$
Initial velocity of second block, $u_2=0m/s$

The momentum before collision is:
$p_i=m_1{u}_1+m_2{u}_2$
$p_i=1\times 10+5\times 0$
$p_i=10kgm/s$

Let the final velocity of the blocks be $v$.
Final momentum will be:
$p_f=(m_1+m_2)v$
$p_f=(1+5)v=6v$

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,
$p_f=p_i$
$6v=10$
$v=1.67m/s$