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…

Question

A block of mass m = 10 &#160; 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 &#160; g moving with speed v , that gets embedded in it, the block moves and comes to stop after moving a distance of 2 &#160; 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 &#160; m &#160; s - 2 , the value of H is close to

Aicharya · 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

$W_{friction} = ∆ k$

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

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

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

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

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

$v^{2} = {(201)}^{2} 2$

$v = \sqrt{2} \times 201$

For a freely falling body,

$v' = \sqrt{2 g H}$

$\frac{v}{10} = \sqrt{2 g H}$

$\frac{\sqrt{2} \times 201}{10} = \sqrt{2} \times \sqrt{g H}$

$\frac{{(201)}^{2}}{100} = 10 H$

$H = \frac{{(201)}^{2}}{1000} = 40 m$

$= 0.04 km$

A block of mass m = 10   kg rests on a horizontal table. The coefficient of friction between the block…

Solution

A block of mass m = 10 &#160; kg rests on a horizontal table. The coefficient of friction between the block…

Solution

A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block…