Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

A bullet of mass $10 g$ moving horizontally at a speed of $50 \sqrt{7} m s^{- 1}$ strikes a block of mass $490 g$ kept on a frictionless track as shown in figure. The bullet remains inside the block and the system proceeds towards the semicircular track of radius $0.2 m$ . Where will the block strike the horizontal part after leaving the semicircular track?

Accepted Answer

Given:

Mass of the bullet $= 10 g$

Speed of the bullet $= 50 \sqrt{7} m s^{- 1}$

Mass of the block $= 490 g$

Radius of the track $(r) = 0.2 m$

Since the net external force $F_{net} = 0$ on the system, when the bullet strikes the block.

Therefore, the linear momentum remains conserved just before and after the strike.

In horizontal direction, initial momentum $=$ final momentum

$\Rightarrow (0.01) (50 \sqrt{7}) = (0.01 + 0.49) v_{1}$

$\Rightarrow v_{1} = \frac{0.01 \times 50 \sqrt{7}}{0.5}$

Speed of the block after collision, $v_{1} = \sqrt{7} m s^{- 1}$

At the time when the block leaves the semicircular track, normal contact force becomes zero at that instant.

$∴ m g sinθ = \frac{m v^{^{2}}}{r}$

$\Rightarrow v^{2} = r g sinθ \dots (i)$

Also, the total mechanical energy will remain conserved.

So, change in $K E =$ Work done on the system

$\Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m {v_{1}}^{2} = - m g h$

$\Rightarrow v^{2} - {v_{1}}^{2} = - 2 g (r + r sinθ)$

$\Rightarrow g r sinθ - 7 = - 2 g r - 2 g r sinθ$

$\Rightarrow g r (3 sinθ + 2) = 7$

$\Rightarrow 3 sinθ = \frac{7}{10 \times 0.2} - 2 = \frac{7}{2} - 2$

$\Rightarrow 3 sinθ = \frac{3}{2}$

$\Rightarrow sinθ = \frac{1}{2} \Rightarrow θ = 30 °$

From equation $(i)$ , $v = \sqrt{(0.2) \times 10 \times \frac{1}{2}} = 1 m s^{- 1}$

Horizontal distance of the block from the centre of the circular track, $d = r cosθ = \frac{\sqrt{3}}{10} m$

After losing contact with the circular track, the block falls back by following projectile path as shown in the figure,

From the figure, $h = r + r sinθ = 0.2 + 0.2 (\frac{1}{2}) = 0.3 m$

Time taken by the block to reach the horizontal surface(ground),

from $s = u t + \frac{1}{2} a t^{2}$

$\Rightarrow - h = v cosθ t + \frac{1}{2} (- g) t^{2}$

$t = \frac{v cosθ + \sqrt{{(v cosθ)}^{2} + 2 g h}}{g}$

$t = \frac{1 (\frac{\sqrt{3}}{2}) + \sqrt{{(\frac{\sqrt{3}}{2})}^{2} + 2 (10) (0.3)}}{10}$

$t = \frac{\sqrt{3}}{5} s$

Horizontal distance from the point of leaving, where the block hit the track, $l = [v sinθ] \times t$

$∴ l = 1 \times \frac{1}{2} \times \frac{\sqrt{3}}{5} = \frac{\sqrt{3}}{10} m$

It is exactly equal to the distance of the block at the time of leaving the track from the centre of the circular track.

It means, the block falls on the track just below the centre of circular track, i.e., at point where circular part of track joins the horizontal part of the track.

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