Question

A bullet of mass 0.012 kg and horizontal speed 70 $m{{s}^{-1}}$ strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

Hint: Check whether the net force is zero to use the law of conservation of momentum before and after collision. Also use the work-energy theorem to find the height to which the block rises. To find the heat produced, use the concept of conservation of energy.

Formula used:
p=mv
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
It is given that the bullet instantly comes to rest with respect to the block. This means that the bullet and the block do not have a relative motion, which means that they are travelling at the same velocities.

Therefore, the bullet of mass 0.012 kg comes with a horizontal speed 70$m{{s}^{-1}}$ and hits the block of mass 0.4 kg, suspended from the ceiling and after this instant they both travel together at the same velocity.

Consider the bullet and the block as one system.

Now, let us talk about the forces acting on the system. There are only two forces acting on the system. The gravitational force, which is equal to mg, acting on the block and the tension in the wire by which the block is suspended.

Here, the tension (T) is balancing the force of gravity. Therefore, mg = T.
Let us assume that the gravitational force on the bullet is negligible.

Hence, the net force on the system will be zero.

If the net force on the system is zero, there will be conservation of momentum of the system before and after the collision.

Momentum (p) is the product of the mass and the velocity of a body.

Therefore, momentum before the collision is ${{p}_{1}}=0.012kg\times 70m{{s}^{-1}}=0.84kgm{{s}^{-1}}$.

After the collision, the bullet and the block travel together. We can consider both of them a single body of mass 0.412 kg. Let the velocity of this body be v.

Therefore, momentum of the system after the collision is ${{p}_{2}}=0.412vkgm{{s}^{-1}}$.

We know that ${{p}_{1}}={{p}_{2}}$.
Hence, $0.84kgm{{s}^{-1}}=0.412vkgm{{s}^{-1}}$
$v=\dfrac{0.84}{0.412}=2.04m{{s}^{-1}}$.

This means that the bullet and the block are moving together at a speed of $2.04m{{s}^{-1}}$.

After the collision, the system will move in a circular motion and the gravitational force will affect its motion. Due to this, it will reach a certain height and stop. Let that height be h.

The system comes to rest because the gravitational force does a negative work on the system, due to which its kinetic energy is brought to zero.

That means the change is kinetic energy is equal to the work done on the system.

Here, the change in kinetic energy is $0-\dfrac{1}{2}m{{v}^{2}}=0-\dfrac{1}{2}(0.412){{\left( 2.04 \right)}^{2}}=-\dfrac{1}{2}(0.412){{\left( 2.04 \right)}^{2}}$ .
Work done on the system is -mgh = -(0.412)(9.8)h.

Hence,

$-\dfrac{1}{2}(0.412){{\left( 2.04 \right)}^{2}}=-\left( 0.412 \right)\left( 9.8 \right)h$
$\Rightarrow h=\dfrac{\dfrac{1}{2}(0.412){{\left( 2.04 \right)}^{2}}}{\left( 0.412 \right)\left( 9.8 \right)}=0.212m$.

Therefore, the block rises to a height of 0.212 metres.

Let us calculate the heat produced during this process.

Since it is an inelastic collision, there will be some Amount of kinetic energy converted to heat energy.

Heat produced will be equal to the difference in initial and final kinetic energy of the system.

Hence, heat produced = $\dfrac{1}{2}(0.012){{(70)}^{2}}-\dfrac{1}{2}(0.412){{(2.04)}^{2}}=29.4-0.86=28.54J$

Note: We know that energy can neither be created nor destroyed. It can only be converted from one form of energy to another. In the above case, the bullet comes with some amount of kinetic energy and hits the block. During the inelastic collision, some amount of this kinetic energy is converted into heat and the rest of the energy is in the form of the kinetic energy of the block and the bullet moving together.
Hence, we can write the following equation,
${{K}_{initial}}={{K}_{final}}+heat$