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course Phy 201
022. query 22 submitted 25 Mar 11 around 10:30 AM.
022. query 22
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Question: `qQuery gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west
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Your solution:
Given:
m = 95 kg
v0 = 4 m/s East
dt = .75 s (West)
The magnitude of the momentum P of the fullback is the product of the mass (m) of the fullback and his velocity (v).
p = mv
Substitute 95 kg for m and 4.0 m/s for v.
p =15kg (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.
(b)
The impulse exerted on the fullback is equal to its total change in momentum dp. Let p(final) be the final momentum of the fullback, and let p (initial) be its initial momentum. Therefore,
Impulse = dp
Impulse = p(final) - p(initial)
The initial momentum of the fullback is 380 kg m/s, and since he comes to rest, the final momentum would be zero. Substitute 0 for p(final) and 380 kg m/s for p(initial). Find the impulse.
Impulse = 0 kg m/s - 380 kg m/s = -380 kg m/s
(c)
Choose the x-axis to be positive towards the right. Therefore, the impulse exerted on the tackler and the impulse exerted on the fullback by the tackler would be equal in magnitude but opposite sign. Impulse is negative so the direction is in the negative x direction, i.e., West. Thus, the impulse exerted on the tackler is:
(-380 kg m/s)
(d)
The magnitude of the impulse exerted on the tackler is the product of the average force F and the time dt over which the force acts.
Impulse = Fave * `dt
Fave = impulse / dt
Fave = dp / dt
Fave = -380 kg m/s /(.75s) = -506 N
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 impulse on the tackler is to the East. **
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Given Solution:
`a** We'll take East to be the positive direction.
The original 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 impulse on the tackler is to the East. **
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&#Very good responses. Let me know if you have questions. &#