A 125 kg fullback is running at 5.0 m/s to the east and is stopped in 0.75 s by a head-on tackle by a tackler running due west.
a) Calculate the original momentum of the fullback.
i did this and got 625 kg- m/s
b) Calculate the impulse exerted on the fullback
i did this and got 625
(c) Calculate the impulse exerted on the tackler.
I did this and got 625 but D I CANT FIGURE OUT !!!
(d) Calculate the average force exerted on the tackler.
Since the unit for force, a newton is equal to kg*m/s^2 you must divide the momentum, kg*m/s of the player by the time it took for him to be tackled.
To calculate the impulse exerted on the tackler, you can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision since there are no external forces acting on the system.
The original momentum of the fullback is calculated to be 625 kg m/s (as you correctly calculated in part a).
The momentum of the tackler can be determined using the equation:
m1v1 + m2v2 = m1v1' + m2v2'
Where:
m1 = mass of the fullback
v1 = initial velocity of the fullback
m2 = mass of the tackler
v2 = initial velocity of the tackler
v1' = final velocity of the fullback (which is 0 m/s since the fullback comes to a stop)
v2' = final velocity of the tackler
Let's assign some variables:
m1 = 125 kg (mass of the fullback)
v1 = 5.0 m/s (initial velocity of the fullback)
m2 = ? (mass of the tackler, unknown)
v2 = -5.0 m/s (initial velocity of the tackler, negative since it's running due west)
v1' = 0 m/s (final velocity of the fullback)
v2' = ? (final velocity of the tackler, unknown)
Using the conservation of momentum equation and plugging in the known values:
(125 kg)(5.0 m/s) + m2(-5.0 m/s) = (125 kg)(0 m/s) + (m2)(v2')
625 kg m/s - 5m2 = 0 kg m/s + v2'm2
Since the final velocity of the tackler is not given, we'll represent it with v2'. We can rearrange the equation to solve for v2':
v2' = (625 kg m/s - 5m2) / m2
Since you already know the impulse exerted on the fullback is 625 Ns, which is equal to the impulse exerted on the tackler, you can set this value equal to v2'm2:
625 Ns = v2'm2
Substituting the equation for v2' into this equation:
625 Ns = [(625 kg m/s - 5m2) / m2] * m2
Now we can solve for m2 by isolating it:
625 Ns = (625 kg m/s - 5m2)
625 Ns - 625 kg m/s = -5m2
-5m2 = 625 kg m/s - 625 Ns
-5m2 = 0 kg m/s - 625 Ns
m2 = (0 kg m/s - 625 Ns) / (-5)
m2 = 125 kg
Therefore, the mass of the tackler is also 125 kg.
Now that we know the mass of the tackler and the impulse exerted on him is also 625 Ns, we can calculate the average force exerted on him. The impulse exerted on an object is equal to the force multiplied by the time over which the force is exerted:
Impulse = Force * Time
Rearranging the equation to solve for Force:
Force = Impulse / Time
Using the values you calculated:
Force = 625 Ns / 0.75 s
Force = 833.33 N
Therefore, the average force exerted on the tackler is approximately 833.33 Newtons.
d) force*time=changeinmomentum
since you don't have time of impact, you cant do it.
other answers: momentum and impulse are vectors, you didn't indicate direction.