A ball of mass 0.315 kg that is moving with a speed of 5.9 m/s collides head-on and elastically with another ball initially at rest.?

[匿名]

2016-07-12 5:33 am

Immediately after the collision, the incoming ball bounces backward with a speed of 3.6 m/s.

a) Calculate the velocity of the target ball after the collision.
Express your answer to two significant figures and include the appropriate units.

b) Calculate the mass of the target ball.
Express your answer to two significant figures and include the appropriate units.

回答 (1)

micatkie

2016-07-12 5:53 am

✔ 最佳答案

(a)
Take all quantities with the same direction of the initial motion of the incoming ball as position.

Conservation of momentum :
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.315×5.9 + 0 = 0.315×(-3.6) + m₂v₂
m₂v₂ = 0.315×9.5 ...... [1]

Conservation of K.E. for elastic collosion :
(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂²
Then, m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
0.315×5.9² + 0 = 0.315×(-3.6)² + m₂v₂²
m₂v₂² = 0.315×5.9² - 0.315×(-3.6)²
m₂v₂² = 0.315×(5.9² - 3.6²)
m₂v₂² = 0.315×(5.9 + 3.6)(5.9 - 3.6)
m₂v₂² = 0.315×9.5×2.3 ...... [2]

[2] / [1]
v₂ = 2.3
Velocity of the target ball = 2.3 m/s
along the same direction of initial motion of the incoming ball.

(b)
Put v₂ = 2.3 into [1] :
m₂×2.3 = 0.315×9.5
m₂ = 1.3
Mass of the target ball = 1.3 kg

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