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1. A 14 N force is applied for 0.33 seconds. Calculate the impulse.
2. A 48 kg object is moving 12 m/s down. It hits the ground and bounces up moving 10 m/s.
A. Which velocity is negative?
B. Calculate the initial momentum of the object.
48kg
10m/s
12m/s
48kg
C. Calculate the final momentum of the object.
D. Remembering that "change of" is always final – initial,
what is the change of momentum of the object?

1 A 14 N force is applied for 033 seconds Calculate the impulse 2 A 48 kg object is moving 12 ms down It hits the ground and bounces up moving 10 ms A Which vel class=

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Answer:

1. The impulse of the force is 4.62 kg¡m/s

2. A. The negative velocity is the return 10 m/s velocity on the way upwards

B. The initial momentum of the object is 576 kg¡m/s

C. The final momentum of the object is -480 kg¡m/s

D. The change of momentum of the object is -1,056 kg¡m/s

Explanation:

1. The magnitude of the applied force, F = 14 N

The duration of the applied force, Δt = 0.33 seconds

The impulse of the force, I = F × Δt

By substitution of the values, we have;

The impulse of the force, I = 14 N × 0.33 s = 4.62 N·s = 4.62 kg·m/s

2. The mass of the object, m = 48 kg

The initial velocity of the object, v₁ = 12 m/s

The final velocity of the object, v₂ = -10 m/s

A. The initial velocity = 12 m/s (downwards)

The negative velocity is the velocity that is in the opposite direction to the initial downward velocity, therefore, the upwards 10 m/s velocity is the negative velocity

B. The initial momentum of the object, [tex]P_i[/tex] = m₁·v₁

By substituting the values, we have;

[tex]P_i[/tex] = 48 kg × 12 m/s = 576 kg·m/s

C. The final momentum of the object, [tex]P_f[/tex] = m₂·v₂

By substituting the values, we have;

[tex]P_f[/tex] = 48 kg × (-10 m/s) = -480 kg·m/s

D. The change of momentum of the object, ΔP = [tex]P_f[/tex] - [tex]P_i[/tex]

∴ ΔP = -480 kg·m/s - 576 kg·m/s = -1,056 kg·m/s

The change of momentum of the object, ΔP = -1,056 kg·m/s.

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