Question

Part A

Two 78.5-kg hockey players skating at 5.97 m/scollide and stick together.I

f the angle between their initial directions was 140 ∘, what is their speed after the collision? vf= m/s

Part B

A 0.30-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it is 260 N.

If the ball's speed after being struck is 5.1 m/s, for what amount of time was the mallet in contact with the ball? T= ms

Answer 1

Part A )

This is an inelastic collision where momentum is always conserved.
m1*v1 + m2*v2 = (m1+ m2)*vf

Since their masses are equal:
m*(v1 + v2) = 2m*vf

So vf = (v1 + v2) / 2

Let v1 = 5.97 @ 0 deg = 5.97i
so v2 = 5.97 @ 140 deg = -4.573285325i + 3.83744203j

v1 + v2 = 1.396714675i + 3.83744203j = 4.083720512 @ 70 deg

So the final speed is |v1 + v2| / 2 = 4.083720512 m/s / 2 = 2.041860256 m/s

Part B)

mass = 0.3 kg
Force = 260 N = 260 kg-m/s^2

Assuming Uniform force and uniform acceleration
force = mass x acceleration
or f = m*a

therefore a = f/m = 260/0.3 m/s^2
or acceleration = 866.6666667 m/s^2
now initial velocity = 0 (the ball was at rest)
Let final velocity be v, and time of contact be t
Then v = a*t
or time of contact = velocity/acceleration = 5.1 m/s / (866.666667 m/s^2)
or t = 5.884615385*10^-3 sec

The mallet was in contact with the ball for 5.884615385 milli seconds