A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0 . 05 . When hit by a bullet of mass 50 g moving with speed v , that gets embedde

Question

A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0 . 05 . When hit by a bullet of mass 50 g moving with speed v , that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v 10 after being dropped from height H , then neglecting energy losses and taking g = 10 m s - 2 , the value of H is close to

TestHub · Accepted Answer

By the conservation of linear momentum,

m_{1} v = (m + m_{1}) v_{1}

\Rightarrow v_{1} = \frac{m_{1} v}{m + m_{1}} . . . (1)

After collision by work energy theorem, we have

W_{friction} = ∆ k

\Rightarrow - μ (m + m_{1}) g x = - \frac{1}{2} (m + m_{1}) v_{1}^{2}

\Rightarrow μ g x = \frac{1}{2} v_{1}^{2}

\Rightarrow 0.05 \times 10 \times 2 = \frac{1}{2} {(\frac{m_{1} v}{m + m_{1}})}^{2}

\Rightarrow \frac{5}{100} \times 10 \times 4 = {(\frac{50}{1000})}^{2} \times \frac{v^{2}}{{(10 + \frac{50}{1000})}^{2}}

\Rightarrow 2 = \frac{1}{400} \frac{v^{2} \times 400}{{(201)}^{2}}

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