Question

A ball of mass 0.50 kg falls and hits the floor at 10 m/s. It rebounds at speed 8.0 m/s.
The collision between the ball and the floor lasts for 0.50 s.
What is the average force acting on the ball during the collision?
А 2.0 N upwards
B 2.0 N downwards
C 18 N upwards
D 18 N downwards​

Answer 1

Answer:

3.5N/s square

Explanation:

A ball of mass 0.005kg falls and hits the floor at 10m/s it rebounds at speed 8m/s the collision between the ball and the floor lasts for 0.50 seconds what is the average force acting on the ball during collision

3.5N/s square

Answer 2

Ans: C 18N upwards

Explanation:

Step 1: looking at the values given this answer requires us to use momentum and impulse, first we get the change in momentum (or impulse) which is m(v-u) = Ft

Step 2: because it states the ball is REBOUND we know that the ball moved in the opposite direction than the initial, thus being a negative value of the initial, hence instead of 0.5(10-8), we need to use 0l5(10--8) which is the same as 0.5(10+8) so our change in momentum and impulse is 9N/s

Step 3: since we have the time, we know that F*0.5 = 9N/s,  dividing 9\0.5, we find our force to be 18N

Step 4: Now we need to find the direction of force, because we know force has the ability to change direction, the force must have been exerted in the new direction it travels in, hence UPWARDS as it was rebound, So the answer then becomes 18N upwards

Hope this was helpful!