a bullet of mass 10g travels horizontally with a speed of 100m/s and is absorbed by a wooden block of mass 990g suspended by a string. Find the vertical height through which the block rises

 

Answers (1)

 

The mass of the bullet,m1 =10g = 0.010 kg

Mass of the block. m2 =990g =0.990kg

Initial velocity of the bullet, u1 =100 m/s

Initial velocity of the block, u2  = 0

Let v be the velocity acquired by the block and bullet together after collision.

Then according to the principle of conservation of momentum:


\begin{array}{l} \mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v} \\ \mathrm{v}=\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)} \\ =\frac{0.010 \times 100+0.990 \times 0}{(0.010+0.990)} \\ =\frac{1}{1}=1 \mathrm{m} / \mathrm{s} \\ \mathrm{v}=1 \mathrm{m} / \mathrm{s} \end{array}

\begin{aligned} &\text { If the bullet and block rise to a height 'h', then: }\\ &\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}=\left(m_{1}+m_{2}\right) g h\\ &\frac{1}{2} \times=(0.010+0.990) \times 10 \times h\\ &0.5=1 \mathrm{Oh}\\ &h=\frac{0.5}{10}=0.05 \mathrm{m}\\ &\mathrm{h}=5 \mathrm{cm} \end{aligned}

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