a fullback having a mass of 9.0x10^1 kg is running at 4.5 m/s [E]. He is hit by a linebacker and 1.2 s later is moving backwards at 1.5 m/s [W].Find:
a) the impulse imparted to the fullback
B) the average force exerted on the fullback by the linebacker
my work:
impulse is p
so p= p2-p2
p=m2v2-m1v1
=m(v2-v1)
=9.0x10^1(1.5 [w] - 4.5 [E])
=-270 kg x m/s [w]
but its all wrong because the answer is 5.4 x 10^2 kg.m/s [w] apparently. Please help!
watch your signs
east defined as +
initial v = +4.5
final v = -1.5
we want final - initial
or
(-1.5 - 4.5)
impulse = change in momentum
90 (-1.5 - 4.5) = -90(6)
= -540 kg m/s or 540 west
F = rate of change of momentum
= -540/1.2 = - 450 N
or 450 N west
To find the impulse imparted to the fullback and the average force exerted on him by the linebacker, we need to apply the principles of momentum and impulse.
a) Impulse is defined as the change in momentum of an object. It can be calculated using the formula:
Impulse (p) = m * (v2 - v1)
where:
m is the mass of the fullback (9.0x10^1 kg)
v2 is the final velocity of the fullback (-1.5 m/s [W], since he is moving backwards)
v1 is the initial velocity of the fullback (4.5 m/s [E])
Plugging in the values:
p = (9.0x10^1 kg) * (1.5 m/s [W] - 4.5 m/s [E])
= (9.0x10^1 kg) * (-3 m/s)
= -270 kg.m/s [W]
It seems like your calculation is correct, but the negative sign indicates that the impulse is in the opposite direction (west) to the initial velocity.
b) The average force exerted on the fullback can be determined by dividing the impulse by the time interval over which it acts:
Average Force = p / Δt
where:
Δt is the time interval (1.2 s)
Plugging in the values:
Average Force = (-270 kg.m/s [W]) / (1.2 s)
= -225 N [W]
Again, the negative sign indicates that the force is in the opposite direction (west) to the initial velocity.
Therefore, using the given information, the impulse imparted to the fullback is -270 kg.m/s [W], and the average force exerted on the fullback by the linebacker is -225 N [W].
Note: Make sure to double-check the calculations and units to avoid any errors.