Can I get some physics help?

Question

A 85-kg fullback is running at 3.1 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west. Calculate the original momentum and impulse exerted of the fullback. Calculate the average force and exertion exerted on the tackler - I got 300 kg m/s for the momentum of the fullback.

A 0.060-kg tennis ball, moving with a speed of 5.66 m/s , has a head-on collision with a 0.085-kg ball initially moving in the same direction at a speed of 3.60 m/s . Assume that the collision is perfectly elastic. Determine the speed of the 0.060-kg ball after the collision. - I understand that the speed is in the direction of the initial velocity  but I am so lost beyond that

A 990-kg sports car collides into the rear end of a 2100-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.7 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact. What was the speed sports car at impact? - I got 81761 for the KE but I'm not sure where to go from there.

? · Accepted Answer

fullback:
a/ original momentum = mass * velocity; you are given both.
I don't see how you get 300 kg·m/s with this data.

b/ Since he is stopped, the impulse is equal to the negative of his original momentum.

c/ force = impulse / time.
You might not need the sign in your answer.

tennis ball:
Conserve momentum (always!) and because the collision is elastic, conserve energy.

For an elastic, head-on collision, we know (from conservation of energy), that
the relative velocity of approach = relative velocity of separation, or
(5.66 - 3.60) m/s = v - u
where v is the post-collision velocity of the other ball
and u is the post-collision velocity of the tennis ball
so
v = u + 2.06m/s

Now conserve momentum:
0.060kg*5.66m/s + 0.085kg*3.60m/s = 0.060kg*u + 0.085kg*(u + 2.06m/s)
which solves to
u = 3.24 m/s ◄

sports car:
friction work = µmgd = 0.80 * (990+2100)kg*9.8m/s²*2.7m = 65 410 J
and this is also the post-collision KE.

KE = ½mv² = p² / 2m
where p is the momentum; so
p² = 2m*KE = 2 * 3090kg * 65410J = 4.04e8 (kg·m/s)²
and so
p = 2.01e4 kg·m/s

Since the combined mass is 3090 kg, the post-collision velocity was
v = p / m = 6.5 m/s

Finally, conserve momentum during the collision:
990kg * V + 2100kg * 0 = 3090kg * 6.5m/s
solves to
V = 20.3 m/s

I don't know what you found the KE of (one car? both cars?), and don't see how you could have arrived at that value.

Hope this helps!

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Can I get some physics help?

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