A glass marble whose mass is 1/10kg falls from a height of 2.5m and rebounds to a height of 1.6m.If the time durinv which they are in contact be one tenth of a second then the avg force between the marble and the floor is
Answers
Answer:
The force will be 1400 N
Explanation:
According to the problem the mass of the marble is 1/10 kg
height from the ground is 2.5 m , rebounds height is 1.6m
At the time of falling the initial velocity = 0,
let final velocity= v m/s,
acceleration due to gravity = 9.8 m/s^ 2 ,
The velocity of the ball :
v ^2 = u ^2 + 2as [ u= initial velocity]
=> v ^2 = 0 + 2×9.8×2.5
=> v = 7 m/s(downward)
Now while going up
initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s 2 , height = 1.6 m
The velocity of the ball
v^2 = u^2 + 2as
=> 0 = u ^2 - 2×9.8×1.6[this time the velocity and acceleration are oppositely directed so we have the negative sign]
=> u =5.6 m/s (upward)
Therefore, the change in velocity = (7- 5.6) m/s
=> Δv = 1.4 m/s
There the Impulse can be calculated by change in linear momentum,
dp = mΔv
= 100 x 1.4=140 kg m/s
Therefore the force= dp/dt = 140/1/10 = 1400 N