Question

A bullet of mass 0.012 kg and horizontal speed 70 m/s 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

Given :

Mass of the bullet, m = 0.012 kg

Initial speed of the bullet, u₁ = 70m/s

Mass of the wooden block, M = 0.4kg

Initial speed of the wooden block, u₂ = zero

To find :

The height to which the block rises and the amount of heat produced in the block.

Solution :

Firstly we have to find the final speed of the system of the bullet and block.

using the law of conversion of momentum that is,

$\leadsto \sf mu_1 + Mu_2 = (m + M)v$

substituting all the given values in the formula,

$\leadsto \sf(0.012)(70) + (0.4)(0) = (0.012 + 0.4)v$

$\leadsto \sf 0.84 + 0 = (0.412)v$

$\leadsto \sf v = \dfrac{0.84}{0.412}$

$\leadsto \sf v = 2.04 \: {ms}^{ - 1}$

Now, using the law of conservation of energy that is potential energy at the highest point is equal to kinetic energy at the lowest point that is,

$\leadsto \sf mgh = \dfrac{1}{2} m {v}^{2}$

According to Question,

Mass of system, m' = m + M = 0.012 + 0.4 = 0.412kg.

substituting all the given values in the formula,

$\leadsto \sf m'gh = \dfrac{1}{2} m' {v}^{2}$

$\leadsto \sf \not m'gh = \dfrac{1}{2} \not m' {v}^{2}$

$\leadsto\sf (9.8)h= \dfrac{1}{2} \times {(2.04)}^{2}$

$\leadsto \sf h= \dfrac{2.04 \times 2.04}{2 \times 9.8}$

$\leadsto \sf h= \dfrac{4.161}{19.6}$

$\leadsto\sf h= 0.2123 \: m$

thus, the wooden block will rise to a height of 0.2123 m.

We know,

Head produced = kinetic energy of the bullet - kinetic energy of the system.

$\leadsto \sf H = \dfrac{1}{2} {mu}^{2} - \dfrac{1}{2} {mv}^{2}$

$\leadsto\sf H = \dfrac{1}{2} \times 0.012 \times (70)^{2} - \dfrac{1}{2} \times 0.412 \times (2.04)^{2}$

$\leadsto \sf H = 29.4 - 0.857$

$\leadsto \sf H = 28.54 \: J$

thus, the amount of heat produced in the block is 28.54 J.