12 exam-style questions on linear momentum, impulse, conservation of momentum and variable forces. Tap Show answer for full working.
A particle of mass \(2\,\mathrm{kg}\) is moving with velocity \(5\,\mathrm{m\,s}^{-1}\). Find its momentum. A constant force then acts for \(3\,\mathrm{s}\), changing the velocity to \(11\,\mathrm{m\,s}^{-1}\) in the same direction. Find the magnitude of the force.
Momentum \(=mv=10\,\mathrm{kg\,m\,s}^{-1}\).
Change in momentum \(=2(11-5)=12\). Impulse \(=Ft=12\Rightarrow F=\dfrac{12}{3}=4\,\mathrm{N}\).
Two particles of masses \(3\,\mathrm{kg}\) and \(5\,\mathrm{kg}\) move towards each other on a smooth horizontal surface with speeds \(4\,\mathrm{m\,s}^{-1}\) and \(2\,\mathrm{m\,s}^{-1}\) respectively. After collision they coalesce. Find their common speed after the collision.
Conservation of momentum (take right as positive for the \(3\,\mathrm{kg}\)): \(3\cdot4+5\cdot(-2)=8v\Rightarrow 12-10=8v\Rightarrow v=\dfrac{1}{4}\,\mathrm{m\,s}^{-1}\) in the original direction of the \(3\,\mathrm{kg}\) particle.
A ball of mass \(0.5\,\mathrm{kg}\) hits a wall perpendicularly with speed \(8\,\mathrm{m\,s}^{-1}\) and rebounds with speed \(6\,\mathrm{m\,s}^{-1}\). Find the magnitude of the impulse exerted by the wall on the ball.
Change in momentum \(=0.5(-6)-0.5(8)=-3-4=-7\). Magnitude of impulse \(=7\,\mathrm{N\,s}\).
A constant force of magnitude \(12\,\mathrm{N}\) acts on a particle of mass \(4\,\mathrm{kg}\) for \(5\,\mathrm{s}\). The particle is initially at rest. Find the final speed of the particle and the distance travelled during the \(5\,\mathrm{s}\).
Impulse \(=Ft=60=mv\Rightarrow v=\dfrac{60}{4}=15\,\mathrm{m\,s}^{-1}\).
Acceleration \(a=\dfrac{12}{4}=3\,\mathrm{m\,s}^{-2}\). Distance \(s=\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot3\cdot25=37.5\,\mathrm{m}\).
A particle of mass \(m\) is moving with velocity \(u\mathbf{i}\) when it receives an impulse \(J\mathbf{j}\). Find the magnitude of the velocity immediately after the impulse and the angle the new direction makes with the original direction.
Change in momentum \(=J\mathbf{j}\), so final momentum \(=mu\mathbf{i}+J\mathbf{j}\). Final velocity \(=u\mathbf{i}+\dfrac{J}{m}\mathbf{j}\).
Speed \(=\sqrt{u^2+(J/m)^2}\). Angle \(\theta=\tan^{-1}\bigl(\dfrac{J}{mu}\bigr)\).
Two particles \(A\) and \(B\) of masses \(2\,\mathrm{kg}\) and \(3\,\mathrm{kg}\) are connected by a light inextensible string. They rest on a smooth horizontal surface. Particle \(A\) is given a horizontal impulse of \(12\,\mathrm{N\,s}\) away from \(B\). Find the speed of each particle immediately after the impulse and the tension impulse in the string.
Let the impulse in the string be \(J\). For \(A\): \(12-J=2v_A\). For \(B\): \(J=3v_B\). Since the string is inextensible, \(v_A=v_B=v\).
\(12-J=2v\), \(J=3v\). Add: \(12=5v\Rightarrow v=2.4\,\mathrm{m\,s}^{-1}\). \(J=7.2\,\mathrm{N\,s}\).
A force \(\mathbf{F}=(3t\mathbf{i}+2\mathbf{j})\,\mathrm{N}\) acts on a particle of mass \(2\,\mathrm{kg}\) for \(4\,\mathrm{s}\), starting from rest at the origin. Find the velocity and position of the particle at \(t=4\).
Acceleration \(\mathbf{a}=\dfrac{1}{2}(3t\mathbf{i}+2\mathbf{j})=\dfrac{3}{2}t\mathbf{i}+\mathbf{j}\).
\(\mathbf{v}=\int\mathbf{a}\,\mathrm{d}t=\dfrac{3}{4}t^2\mathbf{i}+t\mathbf{j}\) (from rest). At \(t=4\): \(\mathbf{v}=12\mathbf{i}+4\mathbf{j}\).
\(\mathbf{r}=\int\mathbf{v}\,\mathrm{d}t=\dfrac{1}{4}t^3\mathbf{i}+\dfrac{1}{2}t^2\mathbf{j}\). At \(t=4\): \(\mathbf{r}=16\mathbf{i}+8\mathbf{j}\).
A bullet of mass \(20\,\mathrm{g}\) is fired into a block of mass \(480\,\mathrm{g}\) which is at rest on a smooth horizontal surface. The bullet embeds itself in the block and the combined system moves with speed \(4\,\mathrm{m\,s}^{-1}\). Find the speed of the bullet just before impact.
Conservation of momentum: \(0.02u=(0.02+0.48)\cdot4=2\). \(u=\dfrac{2}{0.02}=100\,\mathrm{m\,s}^{-1}\).
A particle of mass \(m\) moves under a force \(\mathbf{F}=-k\mathbf{v}\), where \(k\) is a positive constant and \(\mathbf{v}\) is the velocity. Given that the particle starts from the origin with velocity \(u\mathbf{i}\), find the velocity and position as functions of time, and show that the particle approaches a limiting position.
\(m\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=-k\mathbf{v}\Rightarrow\dfrac{\mathrm{d}\mathbf{v}}{\mathbf{v}}=-\dfrac{k}{m}\,\mathrm{d}t\Rightarrow\ln|\mathbf{v}|=-\dfrac{k}{m}t+C\). \(\mathbf{v}=u\mathrm{e}^{-kt/m}\mathbf{i}\).
\(\mathbf{r}=\int\mathbf{v}\,\mathrm{d}t=-\dfrac{mu}{k}\mathrm{e}^{-kt/m}\mathbf{i}+D\). At \(t=0\), \(\mathbf{r}=\mathbf{0}\Rightarrow D=\dfrac{mu}{k}\mathbf{i}\). \(\mathbf{r}=\dfrac{mu}{k}(1-\mathrm{e}^{-kt/m})\mathbf{i}\).
As \(t\to\infty\), \(\mathbf{r}\to\dfrac{mu}{k}\mathbf{i}\).
Three particles \(A\), \(B\) and \(C\) of masses \(m\), \(2m\) and \(3m\) lie at rest in a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\). Particle \(A\) is projected towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(\dfrac{1}{2}\), and between \(B\) and \(C\) is \(\dfrac{1}{3}\). Find the speeds of the particles after both collisions have taken place (assuming \(A\) does not meet \(B\) again).
First collision \(A\)–\(B\): momentum \(mu=mv_A+2mv_B\). Restitution \(\dfrac{1}{2}u=v_B-v_A\). Solving: \(v_A=0\), \(v_B=\dfrac{u}{2}\).
Second collision \(B\)–\(C\): momentum \(2m\cdot\dfrac{u}{2}=2mv_B'+3mv_C\Rightarrow u=2v_B'+3v_C\). Restitution \(\dfrac{1}{3}\cdot\dfrac{u}{2}=v_C-v_B'\). Solving yields \(v_B'=\dfrac{u}{10}\), \(v_C=\dfrac{4u}{15}\) (or exact fractions from simultaneous equations).
A particle of mass \(2\,\mathrm{kg}\) is acted on by a force \(\mathbf{F}=(4t\mathbf{i}+(6-2t)\mathbf{j})\,\mathrm{N}\). At time \(t=0\) the particle is at the origin with velocity \(3\mathbf{i}-\mathbf{j}\). Find the impulse of the force over the interval \(0\le t\le 3\) and the velocity at \(t=3\).
Impulse \(=\int_0^3\mathbf{F}\,\mathrm{d}t=\bigl[2t^2\mathbf{i}+(6t-t^2)\mathbf{j}\bigr]_0^3=18\mathbf{i}+(18-9)\mathbf{j}=18\mathbf{i}+9\mathbf{j}\).
Change in momentum \(=\text{impulse}=18\mathbf{i}+9\mathbf{j}\). Initial momentum \(=2(3\mathbf{i}-\mathbf{j})=6\mathbf{i}-2\mathbf{j}\). Final momentum \(=24\mathbf{i}+7\mathbf{j}\). Velocity \(=12\mathbf{i}+\dfrac{7}{2}\mathbf{j}\).
A truck of mass \(800\,\mathrm{kg}\) is moving at \(5\,\mathrm{m\,s}^{-1}\) when it collides with a stationary truck of mass \(1200\,\mathrm{kg}\). After the collision the two trucks move together. Find the common speed and the loss in kinetic energy.
Momentum: \(800\cdot5=2000v\Rightarrow v=2\,\mathrm{m\,s}^{-1}\).
Initial KE \(=\dfrac{1}{2}\cdot800\cdot25=10000\,\mathrm{J}\). Final KE \(=\dfrac{1}{2}\cdot2000\cdot4=4000\,\mathrm{J}\). Loss \(=6000\,\mathrm{J}\).