course Phy 201 z~ް|{ñܶassignment #022
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21:27:00 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 --> m*v=95kg*4.0m/s=380kgm/s The impulse on the fullback makes the fullback go in the negative direction= -3.8*10^2 kgm/s The impulse on the tackler=+3.8*10^2kgm/s The average force= 3.8*10^2kgm/s/.75s=5.1*10^2kgm/s^2=5.1*10^2N
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21:29:40 ** 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 --> Impulse=Fave*`dt Impulse/`dt= Fave Fave=-380kgm/s/.75s=-506N on the fullback =+506N on the tackler
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