Fig. 1 shows a rescue worker catching a supply crate lowered from a helicopter. The crate has mass 25 kg and moves down at 2.4 m/s just before the worker ca...
Question 1 Report
Fig. 1 shows a rescue worker catching a supply crate lowered from a helicopter. The crate has mass 25 kg and moves down at 2.4 m/s just before the worker catches it. The worker lowers their arms as the crate is brought to rest. This makes the stopping time 0.60 s. The crate does not bounce. The rescue team uses this calculation when choosing gloves and protective equipment. Ignore air resistance during the short catch and take downwards as positive. The diagram includes the velocity before the crate is stopped.
(a) What is the crate's momentum before it is caught? [2]
(b) What is the change in momentum as it is caught? [1]
(c) What average force does the worker exert on the crate? [2]
Answer Details
Downwards is positive. The worker must exert an upward force to remove the crate's downward momentum.
- (a) \[p=mv=25\times2.4=60\text{ kg m/s}\]60 kg m/s downwards. [2]
- (b) \[\Delta p=p_f-p_i=0-60=-60\text{ kg m/s}\]\(-60\text{ kg m/s}\), equivalently a change of 60 kg m/s upwards. [1]
- (c) \[F=\frac{\Delta p}{\Delta t}=\frac{-60}{0.60}=-100\text{ N}\]100 N upwards. [2]
Lowering the arms increases stopping time, so the average force needed for this fixed momentum change is reduced.
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