Derive an Expression for elastic potential energy of a spring
1 answer:
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Answer:
a) v = 0.0496 m / s , b) v = 0.0957 m / s
Explanation:
a) in this case we have an inelastic collision,
To solve these problems, the most important thing is to define the system formed by all the bodies, so that the forces during the crash have been internal and the amount of movement is conserved.
Therefore the system is made up of the ball and the player
initial instant. Before catching the ball
p₀ = m v₁
final moment. After you have grabbed the ball
= (m + M) v
the moment is preserved
po = p_{f}
m v₁ = (m + M) v
v = m / (m + M) v₁
let's calculate
v = 0.405 / (0.405 + 69) 8.50
v = 0.0496 m / s
in the same direction that the ball takes
b) In this case the ball bounces with a speed of V₂ = -7.80 m / s, the negative sign is because it is going in opposite directions
let's write the moment in two times
initial instant. Before the crash
p₀ = m v₁
try end. Right after the crash
p_{f} = m v₂ + M v
the moment is preserved
p₀ = p_{f}
m v₁ = m v₂ + M v
v = m (v₁ -v₂) / M
v = 0.405 /69 (8.50 - (7.80))
v = 0.0957 m / s
in the initial direction of the ball