A bullet with mass $m_{b} = 400 g$ and velocity $v_{b} = 100 m/s$ hits a ballistic pendulum of mass $m_{a} = 9.6 kg$ and lodges into it.It makes pendulum swing up from equilibrium position and rises to height $h$ as shown in figure. Determine the value of $h$ . (Take $g = 10 m / s^{2}$ )

0.2 m

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0.6 m

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0.8 m

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0.4 m

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Solution

The correct option is C $0.8 m$
Before collision

After collision

Applying P.C.L.M
$p_{i} = p_{f}$
$m_{b} v_{b} = (m_{a} + m_{b}) v$
$0.4 \times 100 = (9.6 + 0.4) v \Rightarrow v = 4 m/s$

As both bullet and pendulum rises up to maximum height $h$

Using law of conservation of energy,
Work done by gravity $= Δ K . E$
$\Rightarrow - (m_{a} + m_{b}) g h = (K . E)_{f} - (K . E)_{i}$
$\Rightarrow - 10 \times 10 \times h = 0 - \frac{1}{2} (m_{a} + m_{b}) v^{2}$
$\Rightarrow - 10 \times 10 \times h = 0 - \frac{1}{2} \times 10 \times 4^{2}$
$\Rightarrow h = 0.8 m$

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Similar questions

Q. A bullet with mass

m_{b} = 400 g

and velocity

v_{b} = 100 m/s

hits a ballistic pendulum of mass

m_{a} = 9.6 kg

and lodges into it.It makes pendulum swing up from equilibrium position and rises to height

h

as shown in figure. Determine the value of

h

. (Take

g = 10 m / s^{2}

)

A bullet having some initial velocity $v_{B}$ hits a ballistic pendulum of mass $m_{B}$ and lodges in it. When the bullet hits the pendulum it swings up from the equilibrium position and rises up to height $h$ as shown in figure.

Choose correct option.