Define linear momentum and state its SI unit.
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Linear momentum is the product of mass and velocity, p = mv, where m is mass and v is velocity; SI unit kg m s⁻¹.
Momentum And Its Conservation
A trolley of mass 750 g has a momentum of 4.50 kg m s⁻¹ along a track. Calculate its velocity.
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Rearranging for velocity $$v=\frac{p}{m}$$ Converting the mass $$m=0.750\text{ kg}$$ Calculating $$v=\frac{4.50}{0.750}=6.00{\text{ m s}}^{-1}$$ The velocity is 6.00 m s⁻¹.
State how the momentum of a moving object changes if its velocity is tripled while its mass is unchanged. Justify your answer.
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Momentum becomes three times as large / triples; because momentum is directly proportional to velocity at constant mass.
State the principle of conservation of linear momentum.
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The total linear momentum of a system is constant; provided no external resultant force acts on the system / the system is closed.
A 2.0 kg trolley moving at 3.0 m s⁻¹ collides head-on with a stationary 4.0 kg trolley. The two trolleys lock together on impact. Show that their combined velocity immediately after the collision is 1.0 m s⁻¹.
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$${m}_{1}{u}_{1}+{m}_{2}{u}_{2}=({m}_{1}+{m}_{2})v$$ $$(2.0)(3.0)+(4.0)(0)=(6.0)v$$ $$v=\frac{6.0}{6.0}=1.0{\text{ m s}}^{-1}$$ as required.
A 0.40 kg ball moving east at 5.0 m s⁻¹ collides head-on with a 0.60 kg ball moving west at 2.0 m s⁻¹. After the collision the 0.40 kg ball moves west at 1.0 m s⁻¹. Deduce whether the 0.60 kg ball moves east or west after the collision, and calculate its speed.
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Take east as positive. Total momentum before the collision $$p=(0.40)(5.0)+(0.60)(-2.0)$$ $$p=2.0-1.2=0.80{\text{ kg m s}}^{-1}$$ Applying conservation of momentum $$0.80=(0.40)(-1.0)+(0.60){v}_{2}$$ $${v}_{2}=\frac{0.80+0.40}{0.60}=2.0{\text{ m s}}^{-1}$$ The positive sign means the 0.60 kg ball moves east at 2.0 m s⁻¹.
A rifle of mass 4.0 kg fires a bullet of mass 20 g at a muzzle velocity of 360 m s⁻¹. The rifle was at rest before firing. Calculate the recoil velocity of the rifle.
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Total momentum before firing is zero $${m}_{b}{v}_{b}+{m}_{r}{v}_{r}=0$$ Converting the bullet mass $${m}_{b}=0.020\text{ kg}$$ Calculating the recoil velocity $${v}_{r}=-\frac{{m}_{b}{v}_{b}}{{m}_{r}}=-\frac{(0.020)(360)}{4.0}$$ $${v}_{r}=-1.8{\text{ m s}}^{-1}$$ The recoil velocity is 1.8 m s⁻¹ in the opposite direction to the bullet.
A car of mass 1200 kg travelling at 15 m s⁻¹ collides with a stationary car of mass 800 kg on a straight road. The two cars lock together after the impact. Discuss the motion of the cars after the collision and explain how this collision illustrates the link between Newton’s laws and the conservation of momentum. In your answer you should: • calculate the velocity of the wreckage immediately after the collision • explain how Newton’s third law leads to the conservation of total momentum • state one assumption you have made.
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Applying conservation of momentum $${m}_{1}{u}_{1}=({m}_{1}+{m}_{2})v$$ $$v=\frac{(1200)(15)}{2000}=9.0{\text{ m s}}^{-1}$$ in the original direction of travel. During impact the two cars exert equal and opposite contact forces on each other (Newton’s third law). These forces act for the same contact time, so by $$F=\frac{\Delta p}{t}$$ the impulses are equal and opposite. Therefore the changes in momentum of the two cars are equal and opposite, so the total momentum is unchanged; the system is treated as closed / no external resultant force acts on the cars during the collision. Assumption: friction and air resistance are negligible during the brief collision time / the road surface exerts no horizontal external force during impact.