Question 65: Oscillations

Question

A simple pendulum of length $1 \mathrm{~m}$ has a wooden bob of mass $1 \mathrm{~kg}$. It is struck by a bullet of mass $10^{-2} \mathrm{~kg}$ moving with a speed of $2 	imes 10^2 \mathrm{~ms}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :
$$ p_{	ext{initial}} = m_{	ext{bullet}} 	imes v_{	ext{bullet}} $$
After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $ v' $. The final momentum $ p_{	ext{final}} $ is the combined mass of the bullet and bob times the common velocity :
$$ p_{	ext{final}} = (m_{	ext{bullet}} + m_{	ext{bob}}) 	imes v' $$
According to the principle of conservation of linear momentum,
$$ p_{	ext{initial}} = p_{	ext{final}} $$
$$ m_{	ext{bullet}} 	imes v_{	ext{bullet}} = (m_{	ext{bullet}} + m_{	ext{bob}}) 	imes v' $$
Plugging in the values :
$$ (10^{-2} 	ext{ kg}) 	imes (2 	imes 10^2 	ext{ m/s}) = (10^{-2} 	ext{ kg} + 1 	ext{ kg}) 	imes v' $$
Solving for $ v' $ :
$$ v' = \frac{10^{-2} 	imes 2 	imes 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 	ext{ m/s} $$
After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $ h $ that the bob reaches. Using the principle of conservation of energy :
$$ 	ext{Kinetic Energy (KE)}_{	ext{initial}} = 	ext{Potential Energy (PE)}_{	ext{final}} $$
$$ \frac{1}{2}(m_{	ext{bullet}} + m_{	ext{bob}})v'^2 = (m_{	ext{bullet}} + m_{	ext{bob}})gh $$
Isolating $ h $, we get :
$$ h = \frac{\frac{1}{2}(m_{	ext{bullet}} + m_{	ext{bob}})v'^2}{(m_{	ext{bullet}} + m_{	ext{bob}})g} = \frac{v'^2}{2g} $$
Substituting the values for $ v' $ and $ g $ :
$$ h = \frac{(1.98)^2}{2 	imes 10} = \frac{3.9204}{20} = 0.19602 	ext{ m} $$
Looking at the given options, the result most closely matches Option A :
$$ \boxed{0.20 	ext{ m}} $$
Therefore, the height to which the bob rises before swinging back is approximately $ 0.20 	ext{ m} $.

Answer

$0.20 \mathrm{~m}$

Answer

$0.40 \mathrm{~m}$

Answer

$0.30 \mathrm{~m}$

Answer

$0.35 \mathrm{~m}$

Oscillations

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