a ball is dropped from a tower of height 5 m. it accelerates uniformly at 10m/s .if mass of the ball is 2 kg ,find the momentum transferred by a ball to the ground on striking.
mgh=mv²/2
v=sqrt{2gh)
p=mv =m• sqrt{2gh)
To find the momentum transferred by the ball to the ground when it strikes, we need to calculate the final velocity of the ball just before hitting the ground.
We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given:
Height of the tower (s) = 5 m
Acceleration (a) = 10 m/s²
First, let's calculate the initial velocity (u) of the ball.
When the ball is dropped from rest, its initial velocity is 0 m/s.
Using the equation of motion, we can find the final velocity (v):
v^2 = u^2 + 2as
v^2 = 0^2 + 2 * 10 * 5
v^2 = 100
v = √100
v = 10 m/s
Now that we know the final velocity of the ball just before hitting the ground, we can calculate the momentum (p) transferred to the ground.
Momentum (p) is given by the equation: p = mv, where m is the mass of the ball.
Given:
Mass of the ball (m) = 2 kg
Final velocity (v) = 10 m/s
p = m * v
p = 2 * 10
p = 20 kg·m/s
Therefore, the momentum transferred by the ball to the ground on striking is 20 kg·m/s.