Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

A projectile of mass m is fired with a speed

u = 20 m / s

at an angle

θ = 45 °

from a smooth horizontal field. The coefficient of restitution of collision between the projectile and the field is

e = \frac{1}{2}

. How far from the starting point, does the projectile make its third collision with the field?

Accepted Answer

Given,
Initial speed of projectile is, $u = 20 m s^{- 1}$
Angle of projection is, $θ = 45 °$
Coefficient of restitution is, $e = \frac{1}{2}$

Let us assume,
Vertical component of velocity in first, second and third case is, $u_{y}, v_{1 y} & v_{2 y}$
Horizontal component of velocity in first, second and third case is, $u_{x}, v_{1 x} & v_{2 x}$

We should know
Line of impact will be along the normal to the contact surface and coefficient of restitution is to be calculated along line of impact.
Horizontal component of velocity in all the three cases is $u_{x} = v_{1 x} = v_{2 x} = 20 \cos (45 °) = 10 \sqrt{2} m s^{- 1}$
Coefficient of restitution is given by, $e = \frac{velocity of separation}{velocity of approach} . . . (1)$
As projectile is horizontal, its final vertical component of velocity will be equal and opposite to vertical component of velocity at the time of projection.
Time of flight is given by, $T = \frac{2 \times vertical component of velocity}{g} . . . (2)$
Range is given by, $e = \frac{1}{2}$

First case is shown in below figure,

At this case, range will be given by, $R = \frac{u^{2} \sin (2 θ)}{g}$
$\Rightarrow R = \frac{20^{2} \sin (2 \times 45 °)}{10} = 40 m$
From $(1)$ ,
$\Rightarrow \frac{1}{2} = \frac{v_{1 y}}{u_{y}} = \frac{v_{1 y}}{20 \cos (45 °)}$
$\Rightarrow v_{1 y} = \frac{20 \cos (45 °)}{2} = 5 \sqrt{2} m s^{- 1}$

Projectile for second case is shown in below,

|

From $(2)$ ,
Time of flight is, $T_{1} = \frac{2 v_{1 y}}{g} = \frac{2 \times 5 \sqrt{2}}{10} = \sqrt{2} s$
From $(3)$
Range is given by, $R_{2} = v_{1 x} T_{1} = 20 \cos (45 °) \times \sqrt{2}$
$R_{2} = 20 m$
From $(1)$ ,
$\Rightarrow \frac{1}{2} = \frac{v_{2 y}}{v_{1 y}} = \frac{v_{2 y}}{5 \sqrt{2}}$
$\Rightarrow v_{2 y} = \frac{5}{\sqrt{2}} m s^{- 1}$

Similarly third projectile is shown in below figure,

From $(2)$ ,
Time of flight is, $T_{2} = \frac{2 v_{2 y}}{g} = \frac{2 \times \frac{5}{\sqrt{2}}}{10} = \frac{1}{\sqrt{2}} s$
From $(3)$
Range is, $R_{3} = u_{2 x} T_{2} = 20 \cos (45 °) \times \frac{1}{\sqrt{2}} = 10 m$

Distance from starting point to third collision is, $R = R_{1} + R_{2} + R_{3}$
$\Rightarrow R = 40 + 20 + 10 = 70 m$

Important Questions on Centre of Mass, Momentum and Collisions

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A projectile of mass m is fired with a speed u= 20 m/s at an angle θ = 45° from a smooth horizontal field. The coefficient of restitution of collision between the projectile and the field is e=12 . How far from the starting point, does the projectile make its third collision with the field?

Important Questions on Centre of Mass, Momentum and Collisions

Learn and Prepare for any exam you want !