Question

A sphere of mass 0.3 kg moving with a velocity of 4 m/s collides with another sphere of mass 0.5 kg which is at rest. Assuming the collision to be elastic, their velocities after the impact are ​

Answer 1

Solution:-Mass of sphere 1=0.3kg

Mass of second sphere=0.5kg

Initial velocity of sphere1=4m/s

initial velocity of second sphere=0

First method:-Total initial momentum=0.3×4=1.2kgm/s

let the final velocity of sphere1 be V1 and second be V2

total final momentum=0.3V1+0.5V2

As there is no external force acting,so momentum will remain conserved.

hence we can say,initialp=finalp

1.2=0.3V1+0.5V2

3V1+5V2=12 ..i)

Now,as the collision is elastic,so, coefficient of restitution i.e 'e' will be 1

also,we know that

e=velocity of separation/velocity of approach

or e=(v2-v1)/(u1-u2)

so, using this we can write

1=(V2-V1)/(4-0)

V2=V1+4 ....ii)

putting value of equation ii) in first we get

3(V2-4)+5V2=12

8V2=24

V2=3

putting value of V2 in ii) we get,

V1=3-4=-1

hence the velocities of first and second sphere after collision will be -1m/s and 3m/s respectively.

Answer 2

Topic :-

Collision

Given :-

A sphere of mass 0.3 kg moving with a velocity of 4 m/s collides with another sphere of mass 0.5 kg which is at rest. Assume the collision to be elastic.

To Find :-

Velocities of spheres after the impact.

Concept to be Used :-

Momentum Conservation

Total momentum of colliding bodies remains practically unchanged along line of action during collision.

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

Coefficient of Restitution ( e )

It is defined as the ratio of the impulses of recovery and deformation of either body.

$-e=\dfrac{Impulse\:of\:recovery}{Impulse\:of\:deformation}$

$-e=\dfrac{v_{2}-v_{1}}{u_{2}-u_{1}}$

e = 1, for Elastic Collision

e = 0, for Perfectly Inelastic Collision

0 < e < 1, for Inelastic Collision

Solution :-

It is given that :-

$m_{1}=0.3\:kg$

$m_{2}=0.5\:kg$

$u_{1}=4\:m/s$

$u_{2}=0\:m/s$

We will form two equations through which we can calculate final velocities.

Equation 1

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

Put Values,

$0.3(4) + 0.5(0) = 0.3v_{1}+0.5v_{2}$

$1.2 = 0.3v_{1}+0.5v_{2}$

$12 = 3v_{1}+5v_{2}$

Equation 2

For elastic collision, e = 1

$-1=\dfrac{v_{2}-v_{1}}{0-4}$

$4=v_{2}-v_{1}$

Now, solve these equations,

$3(\:4=v_{2}-v_{1}\:)$

$12=3v_{2}-3v_{1}$

$12 = 3v_{1}+5v_{2}$

Now add both equations,

$3v_{2}-3v_{1} + 3v_{1}+5v_{2}=12+12$

$3v_{2}+5v_{2}=12+12$

$8v_{2}=24$

$v_{2}=3$

Put it in any equation,

$12 = 3v_{1}+5v_{2}$

$12 = 3v_{1}+5(3)$

$3v_{1}=12-15$

$v_{1}=-1$

Answer :-

Velocity of sphere of mass 0.3 kg after collision will be -1 m/s and of sphere of mass 0.5 kg will be 3 m/s.

Abbreviation :-

u = Initial Velocity of sphere mass

v = Final Velocity of sphere mass