Momentum and impulseEdexcel International A Level Maths: Revision notes

The momentum of a particle of mass $m$ moving with velocity $v$ is $mv$. It is a vector in the direction of the velocity, with units kg m s⁻¹ (equivalent to N s). In one-dimensional problems choose a positive direction and give each momentum a sign. A particle moving in the negative direction has negative momentum. Example: $3$ kg at $5$ m s⁻¹ in the positive direction has momentum $+15$; $2$ kg at $4$ m s⁻¹ in the negative direction has momentum $-8$. Total momentum $=7$ kg m s⁻¹.

The impulse of a force $F$ acting for time $t$ is $Ft$ for a constant force, in N s. The impulse-momentum principle states: $\text{Impulse}=mv-mu,$ the change in momentum. It is a vector equation, so use signs for direction. Example: a $0.4$ kg ball hits a wall at $12$ m s⁻¹ and rebounds at $8$ m s⁻¹. Taking away from the wall as positive, $I=0.4(8)-0.4(-12)=8$ N s. If contact lasts $0.05$ s, the average force is $\frac{8}{0.05}=160$ N. The impulse on a body is in the direction of the change in velocity.

When two particles collide directly and there is no external force along the line of motion, the total momentum is unchanged: $m_1u_1+m_2u_2=m_1v_1+m_2v_2.$ The particles may move apart, move together at different speeds, or coalesce (join and move with the same speed $v$): $m_1u_1+m_2u_2=(m_1+m_2)v$. Example: $P$ (4 kg, $6$ m s⁻¹) and $Q$ (8 kg, $2$ m s⁻¹ in the opposite direction) coalesce: $24-16=12v$, so $v=\frac23$ m s⁻¹ in the direction of $P$. You do not need Newton's law of restitution in this course; with one unknown speed, momentum conservation is enough.

In a collision, the impulse that $A$ exerts on $B$ is equal and opposite to the impulse $B$ exerts on $A$ (Newton's third law). So you can find it from the change in momentum of either particle, and the two answers should agree in magnitude. Example: $A$ (3 kg) goes from $5$ to $-1$ m s⁻¹ and $B$ (2 kg) from $-4$ to $5$ m s⁻¹. Impulse on $B=2(5)-2(-4)=18$ N s; impulse on $A=3(-1)-3(5)=-18$ N s. Use this check in exams. The direction of the impulse on a particle is the direction of its change of momentum.

Set up the conservation equation with the unknown (a mass $m$ or a speed $v$), then solve. Example: $A$ (3 kg, $8$ m s⁻¹) hits $B$ ($m$ kg, $2$ m s⁻¹, same direction); afterwards $A$ moves at $4$ and $B$ at $7$. $24+2m=12+7m$ gives $m=2.4$. Impulse on $B=2.4(7-2)=12$ N s. A reversed direction means a negative sign. Check that the answer makes sense: a particle cannot pass through another, so the rear particle cannot move faster than the front one after the collision unless the particles have separated.

After a collision a particle can hit a fixed smooth barrier. The barrier exerts an impulse of $m(v+u)$ if the particle rebounds with speed $v$ from incoming speed $u$. Make a timeline: distances, times and speeds before and after each event. Constant speeds give $s=vt$ between events. Example: $B$ (2.4 kg) leaves the collision at $7$ m s⁻¹, hits a barrier $21$ m away after $3$ s, and rebounds at $5$ m s⁻¹. Impulse $=2.4(5+7)=28.8$ N s. $A$ moves at $4$ m s⁻¹, so has covered $12$ m; the gap of $9$ m closes at $4+5=9$ m s⁻¹, so the second collision is $1$ s after the rebound.