Question

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a high - jumper, having just cleared the bar, lands on an air mattress and comes to rest. had she landed directly on the hard ground, her stopping time would have been much shorter. using the impulse - momentum theorem as your guide, determine which one of the following statements is correct.
the air mattress exerts the same impulse, but a greater net average force, on the high - jumper than does the hard ground.
the air mattress exerts a smaller impulse, and a smaller net average force, on the high - jumper than does the hard ground.
the air mattress exerts a greater impulse, and a greater net average force, on the high - jumper than does the hard ground.
the air mattress exerts the same impulse, but a smaller net average force, on the high - jumper than does the hard ground.
the air mattress exerts a greater impulse, but a smaller net average force, on the high - jumper than does the hard ground.

Explanation:

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem is given by \(J=\Delta p = F_{avg}\Delta t\), where \(J\) is the impulse, \(\Delta p\) is the change in momentum, \(F_{avg}\) is the average force, and \(\Delta t\) is the time interval.

Step2: Analyze the change in momentum

The high - jumper's initial velocity \(v_i\) (just before landing) and final velocity \(v_f = 0\) (comes to rest) are the same whether she lands on the air mattress or the hard ground. So, the change in momentum \(\Delta p=m(v_f - v_i)=-mv_i\) (where \(m\) is the mass of the high - jumper) is the same in both cases. So, the impulse \(J=\Delta p\) is the same.

Step3: Analyze the relationship between force and time

From \(J = F_{avg}\Delta t\), we can express \(F_{avg}=\frac{J}{\Delta t}\). Since \(J\) is the same and \(\Delta t_{mattress}>\Delta t_{ground}\) (stopping time on the air mattress is longer), then \(F_{avg,mattress}

Answer:

The air mattress exerts the same impulse, but a smaller net average force, on the high - jumper than does the hard ground.