Question

An object 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 and just after the impact. Also, calculate the velocity of the combined object.

Answer 1

Hint: The concept of conservation of linear momentum is to be applied here.
Linear momentum is the product of mass and the velocity of the object in which the object is moving. It is a vector quantity.As the motion is one dimensional linear momentum to be conserved in the X direction only.In an isolated system, where no external force is acting the linear momentum of the body is always conserved. This is the law of conservation of linear momentum.

Complete answer:
As the motion is one dimensional linear momentum to be conserved in the X direction only.
The final momentum will be due to the combined velocity of both masses as after collision, both masses move off together.

Mass of the first object $\left( {{m}_{1}} \right)=1kg$

Initial Velocity of first object$\left( {{v}_{1}} \right)=10m{{s}^{-1}}$
Mass of the wooden block $\left( {{m}_{2}} \right)=5kg$
Velocity of the wooden block before collision$\left( {{v}_{2}} \right)=0m{{s}^{-1}}$
Total momentum before collision
$\Rightarrow {{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$
$\Rightarrow 10\left( 1 \right)+5\left( 0 \right)$
$\Rightarrow 10kgm{{s}^{-1}}$
It is given that after collision, the object and the wooden block stick together.
Total mass of the system after collision = ${{m}_{1}}+{{m}_{2}}$
Let combined velocity be $V$
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
${{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=\left( {{m}_{1}}+{{m}_{2}} \right)V$
$\Rightarrow 10=6V$
$V=\dfrac{5}{3}m{{s}^{-1}}$.
The total momentum after collision is $\Rightarrow 6\times \dfrac{5}{3}=10kgm{{s}^{-1}}$
Total momentum just before the impact $\Rightarrow 10kgm{{s}^{-1}}$
Hence, velocity of the combined object after collision $V=\dfrac{5}{3}m{{s}^{-1}}$

Note:
Conservation of linear momentum finds its application in a variety of placesOne of the best examples of this is in rocket propulsion in space,there is no medium in space to exert a force on,the rocket system forms an isolated system with no external force acting.Linear momentum is a vector quantity and it needs to be added considering laws of vector addition.