IB Physics SL A.2.18 Impulse-Momentum Change

Practise equating external impulse to vector momentum change, handling rebounds with signed velocities and using collision time to calculate the average resultant or contact force.

Exam points

calculate Δp = m(v - u) with signed velocities so a rebound adds the two speed magnitudes
equate external impulse to Δp and divide by contact time to find average resultant force
explain how extending stopping time reduces average force for the same momentum change

[Maximum number: 3]

A toy rocket is made from a plastic bottle that contains some water.

Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.

Figure for Question A.2.18—Impulse-momentum change question 1 — IB Physics SL

The air-water mixture is called the propellant.
The variation with time of the vertical velocity of the bottle is shown.

The bottle reaches its highest point at time $T_{1}$ on the graph and returns to the ground at time $T_{2}$ . The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.

The bottle bounces when it returns to the ground.

The mass of the bottle is 27 g and it is in contact with the ground for 85 ms .

Determine the average force exerted by the ground on the bottle. Give your answer to an appropriate number of significant figures.

Attempt to use force = momentum change ÷ time $\ll=\frac{(4.5+10) \times 0.027}{85 \times 10^{-3}}=4.6 \gg$

Force = «4.6 + 0.3» 4.9 «N»

Any answer to 2 sf

Marking guidance:

Accept ± 0.5 velocity values from graph

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A.2.18—Impulse-momentum change question 1