Question

39. 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
1) 4 m/s and 0 ms-1 2-1 m/s and 3 m/s
3) 2 m/s and 2 m/s 4) 4 m/s and 8 ms!​

Answer 1

Answer:

Explanation:

Given:

Mass of the sphere in motion (m₁)=0.3kg

Mass of the sphere in rest (m₂)=0.5 kg

Velocity of the sphere (u₁)=4m/s

u₂ =0m/s ( at rest)

To find:

The velocities after elastic collision.

Solution:

In order to find the answer we need to find two equations,

We can easily form two equations with which we can find the final velocity.

Equation 1:

m₁u₁ + m₂u₂= m₁v₁ + m₂v₂  ( we can get this equation from momentum of conservation)

Substituting the values we get,

0.3(4) + 0.5(0) = 0.3v₁+0.5v₂

1.2 = 0.3v₁+0.5v₂

12 = 3v₁+5v₂   ______(1)

For the 2nd equation,

-e=$\frac{v_2-v_1}{u_2-u_1}$(Coefficient of restitution)

For elastic collision, e = 1

-1= $\frac{v_2-v_1}{0-4}$]

4 = v₂-v₁          ______(2)

Now, let's solve these equations,

3×equation(2)

3(4) = 3( v₂-v₁)

12 = 3v₂-3v₁

12 = 3v₁+5v₂

Now add both equations,

3v₂-3v₁ + 3v₁+5v₂ = 12+12

3v₂+5v₂ = 12+12

8v₂ = 24

v₂ = 3

Put it in any equation,

12 = 3v₁+5v₂

12 = 3v₁+5(3)

v₁ = 12-15

v₁ = -1

(v₁,v₂) = (-1,4)

2)  -1 m/s and 4 m/s

Answer 2

Answer:

-1m/s,3m/s

Explanation:

M1=0.3Kg M2=0.5Kg

U1=4m/s. U2=0

V1=m1-m2/M1+m2×4

V1=0.3-0.5/0.8×4

V1=-0.2/0.2= -1m/s

V2=2×0.3/0.8×4

V2=3m/s

ALL THE BEST