An object of mass 1 kg travelling in a straight line with a velocity of $ 10 \mathrm{~m} \mathrm{~s}^{-1} $ collides with, and sticks to, a stationary wooden block of mass $ 5 \mathrm{~kg} $. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Mass of object, $m=1kg$

Velocity of object, $u_1=10ms^{-1}$

Mass of the wooden block, $M=5kg$

Velocity of wooden block, $u_2=0ms^{-1}$

Find out: The total momentum just before the impact and just after the impact. Also, the velocity of the combined object.

Solution:

We know that formula for momentum is given as-

$p=m\times v$

where, 

$p=momentum$

$m=mass$

$v=velocity$

Momentum before the collision (or initial momentum) of the object and wooden block is given as-

$p_o=m\times u_1=1\times 10=10ms^{-1}$

$p_w=M\times u_2=5\times 0=0ms^{-1}$

Thus, the total momentum before collision is = $p_o+p_w=10+0=10ms^{-1}$

Now, 

It is given that after the collision, the object and the wooden block stick together. Therefore,

The total mass of the combined system= $m+M=1+5=6kg$

Velocity of the combined system = $v$

According to the law of conservation of momentum, we know that-

Total momentum before collision (Initial momentum) = Total momentum after collision (Final momentum)

$mu_1+Mu_2=(m+M)v$

$10=6v$

$v=\frac {10}{6}$

$v=\frac {5}{3}ms^{-1}$

Thus, the combined velocity of the object and wooden block, $v=\frac {5}{3}ms^{-1}$.

As the total momentum before the collision is $10ms^{-1}$, then the total momentum after the collision will also be $10ms^{-1}$ according to the law of conservation.