Further momentum and collisions — WPH14 IAL Physics practice questions · Study-IAL

Further momentum and collisions

WPH14Topic 4.13 questions

Impulse, conservation of momentum in two dimensions, and elastic and inelastic collisions.

Specification points

Understand how to use the equation impulse $= F\Delta t = \Delta p$ (Newton's second law of motion)
CORE PRACTICAL 9: Investigate the relationship between the force exerted on an object and its change of momentum
Understand how to apply conservation of linear momentum to problems in two dimensions

CORE PRACTICAL 10: Use ICT to analyse collisions between small spheres, e.g. ball bearings on a table top

Understand how to determine whether a collision is elastic or inelastic

Be able to derive and use the equation $E_k = \frac{p^2}{2m}$

Get a plain-language explanation, a worked example and the usual mistake.

Original practice questions written in the style of the exam.

Question 13 marks
A stationary radium-226 nucleus decays into a radon-222 nucleus by emitting an alpha particle. The alpha particle moves off at $1.5 \times 10^{7}\,\text{m s}^{-1}$ .

Calculate the recoil speed of the radon nucleus. You may use nucleon numbers in place of masses.
Mark scheme
- Total momentum before $= 0$ , so momentum of radon = momentum of alpha (opposite directions) (1)
- $222\,v = 4 \times 1.5 \times 10^{7}$ (1)
- (1)
Question 25 marks
On a smooth horizontal table, puck A of mass $0.17\,\text{kg}$ moves at $2.0\,\text{m s}^{-1}$ and strikes an identical stationary puck B. After the collision, A moves at $1.2\,\text{m s}^{-1}$ at to its original direction.

v = 2.7 \times 10^{5} {m s}^{- 1}

v = 2.7 \times 10^{5}\,\text{m s}^{-1}

v = 2.7 \times 1 0^{5} m s^{- 1}

−1

30^\circ

(a) Calculate the speed of B after the collision and the angle its path makes with A's original direction.

(b) Determine whether the collision is elastic.

Mark scheme

Conserves momentum along the original direction: $2.0 = 1.2\cos 30^\circ + v_x$ , so $v_x = 0.96\,\text{m s}^{-1}$ (1)
Perpendicular: $0 = 1.2\sin 30^\circ + v_y$ , so $v_y = -0.60\,\text{m s}^{-1}$ (masses cancel) (1)
$v_B = \sqrt{0.96^2 + 0.60^2} = 1.1\,\text{m s}^{-1}$ at on the opposite side of the original direction (1)
KE before $= \tfrac{1}{2}(0.17)(2.0)^2 = 0.34\,\text{J}$ ; KE after (1)
KE after is less than KE before, so the collision is inelastic (1)

vy=

−0.60m s−1

1.1m s−1

32^\circ

= \tfrac{1}{2}(0.17)(1.2^2 + 1.13^2) = 0.23\,\text{J}