Question

A bullet of mass \[50g\] is fired from below into the bob of mass $450g$ of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of $1.8m$ . find the speed of the bullet. Take $g = 10m{s^{ - 2}}$
1. $60m{s^{ - 1}}$
2. $120m{s^{ - 1}}$
3. $50m{s^{ - 1}}$
4. $70m{s^{ - 1}}$

Answer 1

Hint We will use the principle of conservation of linear momentum to find relation between velocity of combination and velocity of bullet. Then we will use a third equation of motion to find the required velocity.

Complete Step by step solution
Given: mass of bullet= $m = 50g$
            Mass of bob= $M = 450g$
Let us assume velocity of bullet be v and velocity of combination of bullet and bob be V.
Now by principle of conservation of linear momentum we have,
$mv = (m + M)V$
Putting value of m and M we get,
$
 V = \dfrac{{(0.05kg)v}}{{(0.05kg + 0.45kg)}} \\
  V = \dfrac{v}{{10}}......(1) \\
 $
Now since, the string becomes loose and the bob will go up with a deceleration of $g = 10m{s^{ - 2}}$.
Therefore, it comes to rest at a height of $1.8m$
Using third equation of motion we get,
${v^2} = {u^2} + 2as$
Here,
$
 u = V \\
 v = 0 \\
 a = - g \\
 s = 1.8m \\
 $
Hence using above values, we get,
$
  0 = {V^2} - 2 \times 10 \times 1.8 \\
    \\
 $
Using equation (1), in above equation we get,
$
  {\left( {\dfrac{v}{{10}}} \right)^2} = 36 \\
  v = 60m{s^{ - 1}} \\
 $
Hence the velocity of bullet is $60m{s^{ - 1}}$

Option (1) is correct.

Note Since the motion of bob after being hit by a bullet is against the gravitational acceleration hence its value is taken negative. This negative acceleration is called acceleration which finally leads to bob at rest after reaching a certain height.