Query 22

course PHY 201

July 14 around 5:40 pm

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

Your solution:

The original momentum of the fullback before the collision is: m * `dv = 95 kg * 4 m/s = 380 kg m/s.

The impulse exerted on the fullback is the Fnet multiplied by the time interval. We know the time interval, but we don’t know the net force, but we can solve for it using the Momentum- Impulse Theorem: m * `dv = Fnet * `dt, so: 380 kg m/s = Fnet * .75 s = 380 kg m/s / .75 s = Fnet. Fnet = 506.67 kg m/s^2, or N. So, the impulse is: 506.67 N * .75 s = 380.0025 kg m/s, but since the impulse is equal and opposite of the momentum, then the impulse will be negative: -380 kg m/s.

The impulse that is exerted on the tackler, would be the same as the impulse that was exerted on the fullback, because the tackler hit the fullback.

Then, the average force exerted on the tackler would be the net force: 506.67 N, which is positive because the force exerted on the tackler by the fullback is in the positive direction.

Confidence rating #$&*: 3

.............................................

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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Self-critique (if necessary):OK

Self-critique rating #$&*

3

"

&#This looks good. Let me know if you have any questions. &#

#$&*

Query 22

course PHY 201

July 14 around 5:40 pm

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

Your solution:

The original momentum of the fullback before the collision is: m * `dv = 95 kg * 4 m/s = 380 kg m/s.

The impulse exerted on the fullback is the Fnet multiplied by the time interval. We know the time interval, but we don’t know the net force, but we can solve for it using the Momentum- Impulse Theorem: m * `dv = Fnet * `dt, so: 380 kg m/s = Fnet * .75 s = 380 kg m/s / .75 s = Fnet. Fnet = 506.67 kg m/s^2, or N. So, the impulse is: 506.67 N * .75 s = 380.0025 kg m/s, but since the impulse is equal and opposite of the momentum, then the impulse will be negative: -380 kg m/s.

The impulse that is exerted on the tackler, would be the same as the impulse that was exerted on the fullback, because the tackler hit the fullback.

Then, the average force exerted on the tackler would be the net force: 506.67 N, which is positive because the force exerted on the tackler by the fullback is in the positive direction.

Confidence rating #$&*: 3

.............................................

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):OK

Self-critique rating #$&*

3

"

&#This looks good. Let me know if you have any questions. &#

#$&*