Answer:
(a) speed=v₁ =3.34 m/s
(b) height=0.569 m
Explanation:
Given Data
Clay ball mass=0.045 kg
Other clay ball mass=0.075 kg
To Find
(a) Speed
(b) Height
Solution
Let
m₁ = mass of the first ball 0.045 kg
m₂ = mass of the two balls together, 0.045 kg + 0.075 kg = 0.120 kg
v₁= velocity of the first ball at the point of collision (to be determined)
v₂ = velocity of the two balls immediately after the collision
d₁ = the height in cm the first ball is dropped from (to be determined)
d₂ = the maximum height the two balls rise to, 8.0 cm
g = acceleration due to gravity 9.80 m/sec²
The kinetic energy of the the balls after the collision is equal to the potential energy at the top of their swing, so
(1/2) × (m₂) × (v₂)² = (m₂) × g × (d₂)
(1/2) × (v₂)² = g × (d₂)
v₂ = √[2×g ×(d₂)]
from the conservation of momentum, we know
(m₁)× (v₁) = (m₂)× (v₂)
(m₁)× (v₁) = (m₂) ×√[2×g ×(d₂)]
(v₁) = [(m₂)/(m₁)]×√[2×g ×(d₂)]
v₁ = (O.120 kg/0.045 kg) ×√(2×9.80 m/s² ×0.08 m)
v₁ =3.34 m/s
As we have find the velocity,Now we can calculate the height the first ball (1/2)× (m₁) × (v₁)² = (m₁)× g× (d₁)
(1/2)× (v₁)² = g×(d₁)
(d₁) = {(1/2) (v1)²} / g
d₁={(1/2) (3.34)²} /9.8
d₁=0.569 m
