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A piece of wood of mass 0.03 kg is dropped from the top of a 100m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity $100 ms^{-1},$ from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the combined system reached above the top of the building before falling below is : $(g=10ms^{-2})$
Option 1)
20 m
Option 2)
30 m
Option 3)
40 m
Option 4)
10 m

Answers (1)

best_answer

Inelastic Collision -

Law of conservation of momentum hold good but kinetic energy is not conserved

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\neq \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}$

$m_{1},m_{2}: masses$

$u_{1},v_{1}: initional \: and \: final\: velocities\: of \: mass \: m_{1}$

$u_{2},v_{2}: initional \: and \: final\: velocities\: of \: mass \: m_{2}$

let t=time taken by particles to collide

$t=\frac{d}{V_{rel}}=\frac{100}{100}=1 sec\\at t=1sec\\\\(V_{wood})_{t-1}=gt=10m/s,(S_{wood})_{t-1}=\frac{1}{2}gt^{2}=5m\\\\(V_{bullet})_{t-1}=V-gt=100-10=90m/s\\\\$

just before collision

$m_{1}$ $\downarrow10m/s$

$m_{2}$ $\uparrow10m/s$

just after collision

$V_{2}\uparrow$ $(m_{1}+m_{2})$

Apply momentum conservation

$M_{1}(-10)+M_{2}(go)=(M_{1}+M_{2})V_{2}\\\\M_{1}=0.03kg\\M_{2}=0.02kg \\V_{2}=30m/s$

now combined system with $V_{2}$ attains $h_{max}$

$h_{max}=\frac{V^{2}}{2g}=\frac{30\times 30}{2\times 10}=45m\\height\: \: above\: \: tower=45-(S_{wood})_{t-1}\\\\=45-5\\\\=40m$

Option 1)

20 m

Option 2)

30 m

Option 3)
40 m
Option 4)

10 m

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