Paper 4 Conservation Laws in Physics Answers

Q1 Testing Momentum Conservation

$p_{i} = (0.500) (0.800) = 0.400 kg m s^{- 1}$ and $p_{f} = (0.500 + 0.300) (0.495) = 0.396 kg m s^{- 1}$ . [3]
Percentage difference relative to $p_{i}$ is $∣0.400 - 0.396∣/0.400 \times 100% = 1.0%$ . [2]
Before: $0.005/0.500 \times 100 + 0.010/0.800 \times 100 = 2.25%$ . After: mass uncertainty is $(0.005 + 0.005) /0.800 \times 100 = 1.25%$ and velocity uncertainty is $0.010/0.495 \times 100 = 2.02%$ , giving approximately $3.27%$ . [3]
Yes. The $1.0%$ difference is smaller than the experimental percentage uncertainties, so the measurements are consistent with momentum conservation; they do not prove exact equality. [2]
$K_{i} = \frac{1}{2} (0.500) (0.800)^{2} = 0.160 J$ and $K_{f} = \frac{1}{2} (0.800) (0.495)^{2} = 0.0980 J$ . The decrease, about $0.062 J$ , is transferred to internal energy, sound and deformation. [3]

Vary the lifted mass/load [1]; determine efficiency [1]; keep motor, supply setting, lift height and winding/spool geometry fixed [1].
Measure motor p.d. $V$ , current $I$ and lift time $t$ so $E_{in} = V I t$ [2]. Measure mass $m$ and vertical height $h$ so useful output is $m g h$ [1]. A joulemeter may replace the electrical measurements.
Use at least five suitably spaced safe loads, repeat timings/readings and average [1]. Table should contain raw $m, h, V, I, t$ and derived $V I t$ , $m g h$ and efficiency, with units [1].
Calculate $η = m g h / (V I t) \times 100%$ [1] and plot efficiency against load or present a clearly labelled comparison [1].
Example limitation: current and speed fluctuate during a lift; use a data logger/joulemeter to integrate input energy [2]. Secure the load and keep hands/feet clear of the falling mass, or clamp the motor securely [1]. Accept other specific matched pairs and relevant precautions.