Define momentum.
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Momentum is the product of mass and velocity; p = mv; the SI unit is kg m s⁻¹; momentum is a vector quantity.
Momentum
A truck of mass 4000 kg travels north at 15 m s⁻¹. A car of mass 1200 kg travels south at 25 m s⁻¹. Determine which vehicle has the greater magnitude of momentum and by how much.
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Taking northward as positive: Finding the momentum of the truck $${p}_{\text{truck}}=mv=4000\times 15$$ $${p}_{\text{truck}}=60 000{\text{ kg m s}}^{-1}$$ Finding the momentum of the car $${p}_{\text{car}}=mv=1200\times 25$$ $${p}_{\text{car}}=30 000{\text{ kg m s}}^{-1}$$ Finding the difference $$\Delta p=60 000-30 000=30 000{\text{ kg m s}}^{-1}$$ The truck has the greater magnitude of momentum; it exceeds the car’s momentum by 30 000 kg m s⁻¹.
Define impulse.
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Impulse is the product of force and the time for which the force acts; impulse = FΔt; it is equal to the change in momentum of the object; the SI unit is N s.
A ball of mass 0.15 kg hits a wall at 20 m s⁻¹ and bounces back at 15 m s⁻¹. The contact time with the wall is 0.010 s. Calculate the average force exerted by the wall on the ball.
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Taking the initial direction of the ball (towards the wall) as positive: Step 1: Finding the change in momentum $$\Delta (mv)=m{v}_{\text{final}}-m{v}_{\text{initial}}$$ $${v}_{\text{initial}}=+20{\text{ m s}}^{-1}$$ $${v}_{\text{final}}=-15{\text{ m s}}^{-1}\text{ (bounces back, so negative)}$$ $$\Delta (mv)=(0.15\times (-15))-(0.15\times 20)$$ $$\Delta (mv)=(-2.25)-(3.0)$$ $$\Delta (mv)=-5.25{\text{ kg m s}}^{-1}$$ Step 2: Finding the average force $$F=\frac{\Delta (mv)}{\Delta t}$$ $$F=\frac{-5.25}{0.010}$$ $$F=-525\text{ N}$$ The magnitude of the average force is 530 N (2 s.f.); the negative sign indicates the force acts away from the wall (in the direction of the bounce-back).
State the principle of the conservation of momentum.
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The total momentum of a system remains constant; provided no resultant external force acts on the system.
A cannon of mass 500 kg fires a cannonball of mass 5.0 kg horizontally at 100 m s⁻¹. Calculate the recoil velocity of the cannon.
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Taking the direction of the cannonball as positive: Before firing, everything is at rest, so: $${p}_{\text{before}}=0$$ Applying conservation of momentum: $${p}_{\text{before}}={p}_{\text{after}}$$ $$0={m}_{\text{ball}}{v}_{\text{ball}}+{m}_{\text{cannon}}{v}_{\text{cannon}}$$ $$0=(5.0\times 100)+(500\times {v}_{\text{cannon}})$$ $$0=500+500 {v}_{\text{cannon}}$$ Rearranging for ${v}_{\text{cannon}}$: $${v}_{\text{cannon}}=\frac{-500}{500}$$ $${v}_{\text{cannon}}=-1.0{\text{ m s}}^{-1}$$ The recoil velocity of the cannon is 1.0 m s⁻¹ in the opposite direction to the cannonball.
Define resultant force in terms of momentum.
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The resultant force acting on an object is the change in momentum per unit time; F = Δp / Δt.
A ball of mass 0.50 kg falls vertically and hits the ground at 8.0 m s⁻¹. It bounces back upward at 6.0 m s⁻¹. The ball is in contact with the ground for 0.040 s. Calculate the average resultant force exerted by the ground on the ball.
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Taking upward as positive: Step 1: Finding the change in momentum $${v}_{\text{initial}}=-8.0{\text{ m s}}^{-1}\text{ (downward, so negative)}$$ $${v}_{\text{final}}=+6.0{\text{ m s}}^{-1}\text{ (upward, so positive)}$$ $$\Delta p=m{v}_{\text{final}}-m{v}_{\text{initial}}$$ $$\Delta p=(0.50\times 6.0)-(0.50\times (-8.0))$$ $$\Delta p=3.0-(-4.0)$$ $$\Delta p=7.0{\text{ kg m s}}^{-1}$$ Step 2: Finding the average resultant force $$F=\frac{\Delta p}{\Delta t}$$ $$F=\frac{7.0}{0.040}$$ $$F=175\text{ N}$$ The average resultant force is 175 N upward, which is approximately 180 N (2 s.f.).