Query 22 

course Phy 201

L¡£Æ°Ñ§ñäíÌŽë²Ø_ƒ¼ÏðZ½·¿»Ø”jùÌassignment #022

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Physics I

07-07-2007

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10:26:04

Query gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west

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RESPONSE -->

For the momentum I took 95kg * 4m/s = 380F and then for the impulse exerted on the fullback I took 380F * .75sec = 285, and then I used the same thing for the impulse exerted on the tackler, and I wasn't sure how to calculate the average force exerted.

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10:29:46

** We'll take East to be the positive direction.

The origional magnitude and direction of the momentum of the fullback is

p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East.

The magnitude and direction of the impulse exerted on the fullback will therefore be

impulse = change in momentum or

impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s.

Impulse is negative so the direction is in the negative x direction, i.e., West.

Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is

Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N

The direction is in the negative x direction, i.e., West.

The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N.

The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0).

The iimpulse on the tackler is to the East. **

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RESPONSE -->

I understand and I got the same answer for the first part which was p = m * v1 = 115kg (4m/s) = 380 kg m/s and that the impulse = change in momentum or

impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s and the Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N The direction is in the negative x direction, i.e., West. The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N

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Good work. Let me know if you have questions. &#