IB Physics HL A.2 Forces and momentum questions

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Question 1

Multiple choiceMedium1 mark

A 3.0 kg box is stationary on a horizontal platform. A person exerts a 10 N downward force on the box. Take g=10 N kg−1g=10\ \text{N kg}^{-1} What is the magnitude of the vertical force that the box exerts on the platform?

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Answer: C. Translational equilibrium of the box gives N−mg−10=0N - mg - 10 = 0 so the platform’s force on the box is N=(3.0)(10)+10=40 NN = (3.0)(10)+10 = 40\ \text{N} By Newton’s third law, the box exerts an equal-magnitude force on the platform.

A: Treats the third-law forces as cancelling, although they act on different objects, giving 0 N0\ \text{N}
B: Balances only the box’s weight, omitting the person’s downward force, giving 30 N30\ \text{N}
D: Counts the person’s downward force twice, giving 30+2(10)=50 N30+2(10)=50\ \text{N}
Remember: sum forces on one object for equilibrium; third-law pairs act on different objects.

Question 2

Multiple choiceHard1 mark

A body of mass MM moves at speed uu along the +x+x-direction and explodes into two pieces of masses αM\alpha M and (1−α)M(1-\alpha)M where 0<α<10<\alpha<1 The piece of mass αM\alpha M moves at speed vv perpendicular to the original motion. The external impulse during the explosion is negligible. What is the speed of the other piece divided by uu

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Answer: C. Total momentum is conserved as a vector. The other piece must have momentum components MuMu and −αMv-\alpha Mv so its speed is

(Mu)2+(αMv)2(1−α)M.\frac{\sqrt{(Mu)^2+(\alpha Mv)^2}}{(1-\alpha)M}.

Dividing by uu gives 1+α2(v/u)21−α\dfrac{\sqrt{1+\alpha^2(v/u)^2}}{1-\alpha}

A: assigns mass αM\alpha M rather than (1−α)M(1-\alpha)M to the other piece.
B: adds the perpendicular momentum components as scalar magnitudes.
D: uses only the xx-component and omits the momentum needed to cancel the first piece’s perpendicular component.
Remember: conserve momentum in each perpendicular direction before finding a resultant speed.

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