4.3 - Further Mechanics

Impulse measures the effect of a resultant force acting for a time interval: $\mathbf{J}=\mathbf{F}_{\mathrm{avg}}\Delta t=\Delta\mathbf{p}=m\mathbf{v}-m\mathbf{u}$ for constant mass. Choose a positive direction before subtracting the momenta.

A $0.20\,\mathrm{kg}$ ball reverses from $+6.0$ to $-4.0\,\mathrm{m\,s^{-1}}$. Its impulse is $0.20(-4.0-6.0)=-2.0\,\mathrm{N\,s}$. If contact lasts $0.050\,\mathrm{s}$, the average force is $-40\,\mathrm{N}$; the minus sign gives its direction.

Investigate whether the impulse delivered to a moving object equals its change in momentum. A low-friction trolley of known mass can pass through light gates while a force sensor and data logger record the horizontal force during the same interaction.

Keep the trolley system and mass fixed, minimise friction, and align the sensor with the motion. Plot impulse against change in momentum: agreement is supported by a straight line through the origin with gradient near one, within uncertainty.

For an isolated system, total momentum is a vector conserved through a collision or explosion. Choose perpendicular $x$ and $y$ axes, resolve every velocity before and after, and apply conservation independently to both components.

Assign signs from the chosen axes. For a velocity $v$ at angle $\theta$ from $+x$, use $v_x=v\cos\theta$ and $v_y=v\sin\theta$, changing the sign for leftward or downward components. Solve the two component equations, then recombine unknown components using $v=\sqrt{v_x^2+v_y^2}$ and $\tan\theta=v_y/v_x$, with the quadrant checked.

An equivalent scaled construction places momentum vectors head-to-tail. The before-collision total and after-collision total must be the same resultant; a closed momentum polygon provides a geometrical check.

Use an overhead video or other ICT record to measure the two-dimensional velocities of small spheres immediately before and after a collision. The camera should be fixed perpendicular to a level table, with a length scale in the plane of motion.

Use a level, low-friction surface; avoid spin and glancing vertical motion; select frames close enough to the collision that external impulses remain small. Repeat collisions with different approach directions and quote uncertainty from position and frame resolution.

An isolated collision conserves total momentum whether it is elastic or inelastic. It is elastic only when total kinetic energy is also conserved within experimental uncertainty; otherwise it is inelastic.

If the total kinetic energy is $0.962\,\mathrm{J}$ before and $0.975\,\mathrm{J}$ after, a small difference may be consistent with an elastic collision once measurement uncertainty is considered. A clear decrease means energy has transferred to deformation, internal energy or sound, so the collision is inelastic.

Begin with the classical definitions $E_k=\tfrac12mv^2$ and $p=mv$. Since $v=p/m$, substitution gives $E_k=\tfrac12m(p/m)^2=p^2/(2m)$. The result is valid for a non-relativistic particle of constant rest mass.

For a particle with $p=3.0\times10^{-20}\,\mathrm{kg\,m\,s^{-1}}$ and $m=5.0\times10^{-27}\,\mathrm{kg}$, $E_k=p^2/(2m)=9.0\times10^{-14}\,\mathrm{J}$. Squaring the momentum also squares its power-of-ten factor.

Angular displacement $\theta$ describes the angle swept from a reference direction. In radians, $\theta=s/r$, where $s$ is arc length and $r$ is radius measured in the same length unit; one complete revolution is $2\pi\,\mathrm{rad}=360^\circ$.

$150^\circ=150\pi/180=5\pi/6\,\mathrm{rad}$. Conversely, $1.20\,\mathrm{rad}=1.20(180/\pi)=68.8^\circ$. For an arc of $0.30\,\mathrm{m}$ on radius $0.20\,\mathrm{m}$, $\theta=1.5\,\mathrm{rad}$.

Angular velocity is the rate of change of angular displacement: $\omega=\Delta\theta/\Delta t$, measured in $\mathrm{rad\,s^{-1}}$. In uniform circular motion every point on a rigid body shares $\omega$, while its tangential speed is $v=\omega r$.

A wheel of radius $0.25\,\mathrm{m}$ rotates with period $0.40\,\mathrm{s}$. Then $\omega=2\pi/0.40=15.7\,\mathrm{rad\,s^{-1}}$ and a point on its rim moves at $v=(15.7)(0.25)=3.9\,\mathrm{m\,s^{-1}}$.

At two nearby positions on a circle, draw equal-length velocity vectors tangent to the path. Place their tails together and form $\Delta\mathbf{v}=\mathbf{v}_2-\mathbf{v}_1$. For a small angular displacement $\Delta\theta$, the velocity triangle is similar to the position triangle, so $\Delta v/v\approx\Delta\theta$; $\Delta\mathbf{v}$ points towards the centre.

Divide by the time interval: $a=\Delta v/\Delta t=v(\Delta\theta/\Delta t)=v\omega$. Since $v=r\omega$, this becomes $a=v^2/r=r\omega^2$. The acceleration direction remains radially inward even though its magnitude is constant in uniform circular motion.

Circular motion requires a resultant force directed towards the centre because the velocity direction is continually changing. 'Centripetal force' names this inward resultant; it is supplied by real forces already acting on the object.

Draw only real forces, choose inward as the radial direction, and add their radial components with signs. At the bottom of a vertical circle, an upward tension or reaction must exceed weight to leave an inward resultant; at the top, weight may contribute towards the centre.

After identifying the real forces, set their resultant component towards the centre equal to $ma$: $\sum F_{\mathrm{in}}=ma=mv^2/r=mr\omega^2$. Choose the form containing the quantities actually known.

A $0.50\,\mathrm{kg}$ object moves at $4.0\,\mathrm{m\,s^{-1}}$ in a horizontal circle of radius $2.0\,\mathrm{m}$. The required inward resultant is $mv^2/r=(0.50)(4.0^2)/2.0=4.0\,\mathrm{N}$. The named real force must supply this resultant.