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A.2 · 20 · A.2 Forces and momentum · impulse, explosions, circular motion (practice)

Exam-style questions: impulse, explosions and circular motion

Four worked exam-style problems, then a mixed quiz on impulse, explosions, and circular motion.

SL+HLHarder32 minFree
Hook

Know the physics, then practise the writing. Work each question yourself before you open the solution.

Question 1 · impulse
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A bat strikes a ball

A force acts on a 0.16 kg ball for 4.0 ms. The force rises uniformly to 800 N and then falls uniformly to zero. The ball was initially stationary. (a) Calculate the impulse. (b) Calculate the speed of the ball afterwards. (c) State the average force on the ball.

WORKED EXAMPLE · SOLUTION

Peak force 800 N, total time 4.0 × 10⁻³ s, triangular force–time graph, m = 0.16 kg, u = 0.

    Mixed quiz

    Check 1 of 8

    The area under a force–time graph represents…

    Check 2 of 8

    A 0.45 kg football is kicked from rest and leaves at 20 m s⁻¹. The contact lasts 0.030 s. The average force is…

    Check 3 of 8

    A 60 kg astronaut at rest throws away a 2.0 kg tool at 3.0 m s⁻¹. The astronaut recoils at…

    Check 4 of 8

    A body moves in a circle at constant speed. The resultant force on it…

    Check 5 of 8

    The radius of a circle is halved while the speed is unchanged. The centripetal acceleration…

    Check 6 of 8

    A satellite orbits the Earth in a circle. Which force provides the centripetal force?

    Check 7 of 8

    A 2.0 kg object moves in a circle of radius 0.50 m at ω = 6.0 rad s⁻¹. The centripetal force is…

    Check 8 of 8

    Why does the centripetal force do no work on a body in uniform circular motion?

    Remember earlier lessons
    From lesson A.2 · 14

    Impulse equals…

    From lesson A.2 · 18

    a = …

    Summary card

    Key points

    • • Impulse = area under F–t = Δp; average force = J / Δt.
    • • Explosion: momenta cancel, KE increases.
    • • Circular motion: a = v²/r = ω²r = 4π²r/T², F = ma towards the centre, no work done.
    • • Vertical circle: T − mg = mv²/r (bottom), T + mg = mv²/r (top).

    Formulas

    • J = FΔt = Δp
    • F = mv² / r = mω²r

    Key terms

    Common errors

    • • Mixing up the sign of the weight at the top and the bottom
    • • Forgetting that KE is not conserved in an explosion
    Checks solved: 0 of 8
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