A ball a moving with a speed of 99m/s collides directly with another identical ball b movingw
Answers
Complete Question:
A ball A moving with a speed of 99ms−1 collides directly with another identical ball B moving with a speed v in the opposite direction, A comes to rest after the collision. If the coefficients of restitution is 0.8, the speed of B before collision, is :
( A ) 81ms−2
( B ) 22.5ms−1
( C ) 90ms−1
( D ) 10ms−1
Answer:
The correct answer is option (D) 10 ms⁻¹
Given:
Two identical balls collide with one another.
Let the mass of the balls be m. (identical here means having the same mass and size).
Initial velocity of ball A, u₁ = 99 ms⁻¹
Initial velocity of ball B, u₂ = -v ms⁻¹
After collision ball A comes to rest.
∴ Final velocity of ball A, v₁ = 0 ms⁻¹
Final velocity of ball B be v₂.
Find:
The speed of B before the collision.
Solution:
According to the conservation of momentum, we have
Total momentum of system before collision = Total momentum of system after collision
mu₁ + mu₂ = mv₁ + mv₂
mu₁ - mv₁ = mv₂ - mu₂
m(u₁ - v₁) = m(v₂ - u₂)
u₁ - v₁ = v₂ - u₂
99 - 0 = v₂ - (-v) [cause direction is opposite]
v₂ = 99 - v
v₂ = (99 - v) ms⁻¹
Now, coefficient of restitution, e = 0.8
But by formula,
e = (v₂ - v₁)/(u₁ + u₂)
0.8 = (99 - v - 0)/(99 + v)
0.8 = (99 - v)/(99 + v)
0.8(99 + v) = 99 - v
79.2 + 0.8v = 99 - v
0.8 v + v = 99 - 79.2
1.8v = 19.8
2v = 20 (approximation)
v = 20/2
v = 10 m/s
Hence, the speed of ball b before collision = v = 10 ms⁻¹.
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