A-level Physics
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    3.4 Momentum — Practise with help

    Work each problem first. Open a hint only if you are stuck, and the full working only to check. For every momentum problem, write the positive direction before you substitute.

    Warm-up: remember

    1. A resultant force of 240 N acts on an object of mass 60 kg. Find the acceleration.
    2. A car of mass 1000 kg travels at 12 m s−112\,\text{m}\,\text{s}^{-1}. Find its kinetic energy.
    3. A ball of mass 5.0 kg is dropped from rest from a height of 3.0 m. Find its speed just before it lands. Take g=9.81 m s−2g = 9.81\,\text{m}\,\text{s}^{-2}.
    Answers
    1. a=24060=4.0 m s−2a = \dfrac{240}{60} = 4.0\,\text{m}\,\text{s}^{-2}.
    2. Ek=12×1000×122=72000 J=72 kJE_k = \tfrac{1}{2} \times 1000 \times 12^2 = 72000\,\text{J} = 72\,\text{kJ}.
    3. v=2gΔh=2×9.81×3.0=7.7 m s−1v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.81 \times 3.0} = 7.7\,\text{m}\,\text{s}^{-1}.

    If question 1 goes wrong, use F=maF = ma. If question 2 goes wrong, use Ek=12mv2E_k = \tfrac{1}{2}mv^2. If question 3 goes wrong, use conservation of energy with mgΔh=12mv2mg\Delta h = \tfrac{1}{2}mv^2.

    Practise with help

    1. A car of mass 1500 kg travels at 20 m s−120\,\text{m}\,\text{s}^{-1} due north. Find its momentum.

    Hint

    Momentum is mass multiplied by velocity. It carries the direction of the velocity.

    Next step

    Substitute into p=mvp = mv and state the direction with the answer.

    Full working p=mv=1500×20=3.0×104 kg m s−1 northp = mv = 1500 \times 20 = 3.0 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1} \ \text{north}

    The momentum is a vector, so the direction is part of the answer.

    2. A force of 250 N acts for 0.40 s on a ball of mass 2.0 kg that is initially at rest. Find the impulse and the final speed of the ball.

    Hint

    The impulse is FΔtF\Delta t, and it equals the change in momentum.

    Next step

    Find the impulse first, then use Δp=mv−0\Delta p = mv - 0 to find the final speed.

    Full working FΔt=250×0.40=100 N sΔp=100 kg m s−1v=1002.0=50 m s−1\begin{aligned} F\Delta t &= 250 \times 0.40 = 100\,\text{N}\,\text{s} \\ \Delta p &= 100\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ v &= \frac{100}{2.0} = 50\,\text{m}\,\text{s}^{-1} \end{aligned}

    The impulse is 100 N s100\,\text{N}\,\text{s} and the final speed is 50 m s−150\,\text{m}\,\text{s}^{-1}.

    3. A ball of mass 0.15 kg hits a wall at 12 m s−112\,\text{m}\,\text{s}^{-1} and rebounds at 8.0 m s−18.0\,\text{m}\,\text{s}^{-1}. Find the change in momentum of the ball.

    Hint

    Take the direction towards the wall as positive. The rebound velocity is negative.

    Next step

    Use Δp=m(v−u)\Delta p = m(v - u) with u=+12u = +12 and v=−8.0v = -8.0.

    Full working Δp=m(v−u)=0.15(−8.0−12)=−3.0 kg m s−1\begin{aligned} \Delta p &= m(v - u) \\ &= 0.15(-8.0 - 12) \\ &= -3.0\,\text{kg}\,\text{m}\,\text{s}^{-1} \end{aligned}

    The change in momentum is 3.0 kg m s−13.0\,\text{kg}\,\text{m}\,\text{s}^{-1} away from the wall. The size of the change is larger than the ball's initial momentum, because the ball reverses direction.

    4. A trolley of mass 3.0 kg moves at 4.0 m s−14.0\,\text{m}\,\text{s}^{-1} towards a second trolley of mass 1.0 kg, which is at rest. The trolleys collide and stick together. Find their common velocity.

    Hint

    The total momentum before the collision equals the total momentum after it.

    Next step

    Write m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v and substitute the masses and the initial velocities.

    Full working 3.0×4.0+1.0×0=(3.0+1.0)v12=4.0vv=3.0 m s−1\begin{aligned} 3.0 \times 4.0 + 1.0 \times 0 &= (3.0 + 1.0)v \\ 12 &= 4.0v \\ v &= 3.0\,\text{m}\,\text{s}^{-1} \end{aligned}

    The trolleys move together at 3.0 m s−13.0\,\text{m}\,\text{s}^{-1} in the original direction of the first trolley.

    5. An object of mass 0.80 kg is at rest and explodes into two pieces. One piece has a mass of 0.30 kg and moves at 4.0 m s−14.0\,\text{m}\,\text{s}^{-1}. Find the velocity of the other piece.

    Hint

    The object is at rest before the explosion, so its total momentum is zero.

    Next step

    Write 0=m1v1+m2v20 = m_1 v_1 + m_2 v_2 and put in the known mass and velocity.

    Full working 0=0.30×4.0+0.50vv=−1.20.50=−2.4 m s−1\begin{aligned} 0 &= 0.30 \times 4.0 + 0.50v \\ v &= -\frac{1.2}{0.50} = -2.4\,\text{m}\,\text{s}^{-1} \end{aligned}

    The other piece moves at 2.4 m s−12.4\,\text{m}\,\text{s}^{-1} in the opposite direction to the first piece. The total momentum stays at zero.

    6. The chapter problem. The car and riders have a mass of 500 kg and reach 29.7 m s−129.7\,\text{m}\,\text{s}^{-1} at the bottom of the first drop. At the end of the ride the brakes bring the car to rest in 3.0 s. Find the momentum at the bottom and the average braking force.

    Hint

    The change in momentum is the final momentum minus the initial momentum. At the end the car is at rest.

    Next step

    Find p=mvp = mv at the bottom, then use F=ΔpΔtF = \dfrac{\Delta p}{\Delta t} with Δt=3.0\Delta t = 3.0 s.

    Full working p=mv=500×29.7=1.49×104 kg m s−1Δp=0−1.49×104=−1.49×104 kg m s−1F=ΔpΔt=−1.49×1043.0=−5.0×103 N\begin{aligned} p &= mv = 500 \times 29.7 = 1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ \Delta p &= 0 - 1.49 \times 10^4 = -1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ F &= \frac{\Delta p}{\Delta t} = \frac{-1.49 \times 10^4}{3.0} = -5.0 \times 10^3\,\text{N} \end{aligned}

    The average braking force is 5.0×103 N5.0 \times 10^3\,\text{N}, about 5000 N, opposite to the motion.

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