Question

A stone is dropped from a height of h1 = 3.8 m above the ground. After it bounces, it only reaches a height h2 = 2.3 m above the ground. The stone has mass m = 0.1965 kg. Randomized Variables h1 = 3.8 m h2 = 2.3 m m = 0.1965 kg show answer No Attempt 33% Part (a) What is the magnitude of the impulse I, in kilogram meters per second, the stone experienced during the bounce?

Answer 1

Answer:

3.0 kg·m/s

Explanation:

We use the equation of motion to determine the velocity with which it hits the ground.

• Initial velocity, u = 0 m/s (It was dropped)
• Distance, s = 3.8 m
• Acceleration, a = g (the acceleration due to gravity)
• Final velocity, v

$v^2 = u^2+2as$

$v^2 = 0^2 + 2g\times3.8 = 7.6g$

$v=\sqrt{7.6g}$

At the rebound, we need to determine the initial velocity.

• Initial velocity, u
• Distance, s = 2.3 m
• Final velocity, v = 0 m/s
• Acceleration, a = -g (the acceleration due to gravity, negative since it is upwards)

$v^2 = u^2+2as$

$0^2 = u^2+2(-g)(2.3)$

$u^2 = 4.6g$

$u=\sqrt{4.6g}$

The impulse, by Newton's second law of motion, is equal to the change in momentum.

$I = Ft = mv-mu = m(v-u)$

In the rebound, u is in opposite direction to v.

$I = m(v-(-u)) = m(v+u)$

$I = 0.1965(\sqrt{7.6g}+\sqrt{4.6g})$

Taking g = 9.8 m/s²

$I = 0.1965(\sqrt{7.6g}+\sqrt{4.6g}) = 3.0\ \text{kg}\cdot\text{m/s}$