FM1-1 Momentum and impulse — coverage pack

2 specification leaves · notes, questions, answers and worked methods

FM1-1.1 · Momentum and impulse. The impulse-momentum principle. The principle of conservation of momentum applied to two spheres colliding directly.

For a particle of mass $m$ moving with velocity $v$ along a line, its momentum is $mv$ ; choose and state a positive direction before assigning signs.
Impulse equals change in momentum: $I=m(v-u)$ . It is also the area under a force-time graph, $I=\int F\,dt$ .
During a direct collision, total momentum is conserved when the external impulse on the colliding system is negligible.
A negative velocity means motion opposite to the chosen positive direction; do not replace signed velocities by speeds inside a momentum equation.

1. Rightwards is positive. A $3\,\text{kg}$ particle changes velocity from $4\,\text{m s}^{-1}$ rightwards to $2\,\text{m s}^{-1}$ leftwards. Find its impulse.[2 marks]
Answer
- $-18\,\text{N s}$ , so the impulse is $18\,\text{N s}$ leftwards
Method: Use $I=m(v-u)$ with $u=4$ and $v=-2$ . Hence $I=3(-2-4)=-18\,\text{N s}$ . The negative sign gives the leftward direction.

1. Two spheres collide head-on and remain together. The first has mass $2\,\text{kg}$ and velocity $7\,\text{m s}^{-1}$ ; the second has mass $3\,\text{kg}$ and velocity $-1\,\text{m s}^{-1}$ . Calculate their common velocity.[3 marks]
Answer
- $\dfrac{11}{5}\,\text{m s}^{-1}$ in the positive direction
Method: Conservation of momentum gives $2(7)+3(-1)=(2+3)v$ . Thus $11=5v$ , so $v=11/5\,\text{m s}^{-1}$ in the positive direction.

1. Sphere $A$ , of mass $m\,\text{kg}$ , moves rightwards at $8\,\text{m s}^{-1}$ behind a $2\,\text{kg}$ sphere $B$ moving rightwards at $2\,\text{m s}^{-1}$ . After their direct collision, the velocities of $A$ and $B$ are respectively $1\,\text{m s}^{-1}$ and $5\,\text{m s}^{-1}$ rightwards. Find $m$ and the impulse on $A$ .[4 marks]
Answer
- $m=\dfrac67\,\text{kg}$
- Impulse on $A=-6\,\text{N s}$ , or $6\,\text{N s}$ leftwards
Method: Conservation of momentum gives $8m+2(2)=m(1)+2(5)$ . Therefore $7m=6$ and $m=6/7\,\text{kg}$ . For $A$ , $I=m(v-u)=(6/7)(1-8)=-6\,\text{N s}$ , so the impulse is leftwards.

Vector momentum is $m\mathbf v$ , so its components are found by multiplying every velocity component by the mass.
In vector form, $\mathbf J=m(\mathbf v-\mathbf u)$ , and hence $\mathbf v=\mathbf u+\mathbf J/m$ .
Resolve a vector impulse into perpendicular components before finding its magnitude or the direction of the resulting velocity.
Do not apply the impulse magnitude separately to both components; the signed vector components, not just $|\mathbf J|$ , determine the change in velocity.

1. A $2\,\text{kg}$ particle changes velocity from $(3\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ to $(5\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . Find the impulse vector and its magnitude.[3 marks]
Answer
- $(4\mathbf i+10\mathbf j)\,\text{N s}$
- $2\sqrt{29}\,\text{N s}$
Method: $\mathbf J=m(\mathbf v-\mathbf u)=2[(5-3)\mathbf i+(1-(-4))\mathbf j]=(4\mathbf i+10\mathbf j)\,\text{N s}$ . Its magnitude is $\sqrt{4^2+10^2}=2\sqrt{29}\,\text{N s}$ .

1. A $4\,\text{kg}$ particle initially has velocity $(2\mathbf i-\mathbf j)\,\text{m s}^{-1}$ . It receives an impulse $(-6\mathbf i+8\mathbf j)\,\text{N s}$ . Determine its new speed and the acute angle its velocity makes above the positive $\mathbf i$ direction.[4 marks]
Answer
- Speed $\dfrac{\sqrt5}{2}\,\text{m s}^{-1}$
- Angle $\tan^{-1}2\approx63.4^\circ$
Method: $\mathbf v=\mathbf u+\mathbf J/m=(2\mathbf i-\mathbf j)+(-1.5\mathbf i+2\mathbf j)=(0.5\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . Hence $|\mathbf v|=\sqrt{0.5^2+1^2}=\sqrt5/2$ , and $\tan\theta=1/0.5=2$ .

1. A $3\,\text{kg}$ particle has initial velocity $(4\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ . An impulse of magnitude $15\,\text{N s}$ leaves its final velocity perpendicular to its initial velocity. Find all possible final velocities and the corresponding impulse vectors.[5 marks]
Answer
- $\mathbf v=(\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ with $\mathbf J=(-9\mathbf i+12\mathbf j)\,\text{N s}$
- $\mathbf v=(-\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ with $\mathbf J=-15\mathbf i\,\text{N s}$
Method: Write $\mathbf v=x\mathbf i+y\mathbf j$ . Perpendicularity gives $4x-2y=0$ , so $y=2x$ . Also $\mathbf J=3(\mathbf v-\mathbf u)=[3(x-4)\mathbf i+(6x+6)\mathbf j]$ . The magnitude condition gives $9(x-4)^2+(6x+6)^2=225$ , which simplifies to $45x^2+180=225$ . Thus $x=\pm1$ . Substitution gives the two stated velocity and impulse pairs.