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Question: A bullet of mass \[50g\] is fired from below into the bob of mass \(450g\) of a long simple pendulum...

A bullet of mass 50g50g is fired from below into the bob of mass 450g450g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8m1.8m . find the speed of the bullet. Take g=10ms−2g = 10m{s^{ - 2}}
1. 60ms−160m{s^{ - 1}}
2. 120ms−1120m{s^{ - 1}}
3. 50ms−150m{s^{ - 1}}
4. 70ms−170m{s^{ - 1}}

Explanation

Solution

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=50gm = 50g
Mass of bob= M=450gM = 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)Vmv = (m + M)V
Putting value of m and M we get,
V=(0.05kg)v(0.05kg+0.45kg) V=v10......(1)  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=10ms−2g = 10m{s^{ - 2}}.
Therefore, it comes to rest at a height of 1.8m1.8m
Using third equation of motion we get,
v2=u2+2as{v^2} = {u^2} + 2as
Here,
u=V v=0 a=−g s=1.8m  u = V \\\ v = 0 \\\ a = - g \\\ s = 1.8m \\\
Hence using above values, we get,
0=V2−2×10×1.8   0 = {V^2} - 2 \times 10 \times 1.8 \\\ \\\
Using equation (1), in above equation we get,
(v10)2=36 v=60ms−1  {\left( {\dfrac{v}{{10}}} \right)^2} = 36 \\\ v = 60m{s^{ - 1}} \\\
Hence the velocity of bullet is 60ms−160m{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.