Answer:
The magnitude of the force exerted by the ball on the ground during the 0.5 s of contact = 90.16 N
Explanation:
Given,
Mass of mud ball = m = 2.40 kg
Height the ball is released from = y = 18 m
Total contact time of ball and the ground = t = 0.5 s
The Newton's second law of motion explains that the magnitude of the change of momentum is equal to the magnitude of a body's impulse.
Change in momentum = Magnitude of Impulse
Change in momentum = (final momentum) - (initial momentum)
Since the ball is dropped from rest, initial momentum = 0 kgm/s
But to calculate its final momentum, we need the ball's final velocity before hitting the ground.
Using the equations of motion,
u = initial velocity of the ball = 0 m/s (ball was dropped from rest)
v = final velocity of the ball = ?
g = acceleration due to gravity = 9.8 m/s²
y = vertical distance covered by the ball = 18 m
v² = u² + 2gy
v² = 0² + (2)(9.8)(18)
v² = 352.8
v = 18.78 m/s
Final momentum of the ball = (m)(v)
= (2.4) × (18.78) = 45.08 kgm/s
Change in momentum = 45.08 - 0 = 45.08 kgm/s
Impulse = Ft
Change in momentum = Magnitude of Impulse
45.08 = F × (0.5)
F = (45.08/0.5) = 90.16 N
Hope this Helps!!!
Answer:
90.14 N
Explanation:
according to the impulse momentum theorem,
Impulse = change in momentum
Where impulse = force × time and change in momentum = m ( v - u).
The object was initially at rest, hence it initial velocity is zero.
To get the final velocity, we use the formula below
v² = u² + 2gh
Where h = height of the cliff = 18.0m
v² = 2 × 9.8 × 18
v² = 352.8
v = √333.2
v = 18.78 m/s
At t = 0.50s and v = 18.78 m/s, we can get the average force of impact
F×0.50 = 2.4 (18.78 - 0)
F × 0.50 = 2.4 (18.78)
F × 0.50 = 45.072
F = 45.072 /0.50
= 90.14 N
