A ball of 0.5kg mass hits a wall at an angle of 30° and is then reflected making an angle of 90° with its original direction. if the duration of contact between the ball and the wall is 0.1s,calculate the force exerted on the wall. the initial and final velocities of ball are 10ms-1 and 20ms-1.

To calculate the force exerted on the wall by the ball, we'll need to use the principle of impulse and momentum.

Let's break down the problem step by step:

1. Calculating the initial and final momentum of the ball:
The momentum (p) of an object is given by the formula p = m * v, where m is the mass and v is the velocity.

Given:
- Mass of the ball (m) = 0.5 kg
- Initial velocity of the ball (v1) = 10 m/s
- Final velocity of the ball (v2) = 20 m/s

Therefore, the initial momentum (p1) of the ball is given by:
p1 = m * v1 = 0.5 kg * 10 m/s = 5 kg m/s

And the final momentum (p2) of the ball is given by:
p2 = m * v2 = 0.5 kg * 20 m/s = 10 kg m/s

2. Calculating the change in momentum:
The change in momentum (∆p) of the ball is given by the formula ∆p = p2 - p1.

Therefore, the change in momentum of the ball is:
∆p = p2 - p1 = 10 kg m/s - 5 kg m/s = 5 kg m/s

3. Calculating the force exerted on the wall:
The force exerted on the ball by the wall is equal to the rate of change of momentum, which can be calculated using the formula F = ∆p / ∆t, where ∆t is the duration of contact between the ball and the wall.

Given:
- Duration of contact (∆t) = 0.1 s

Therefore, the force exerted on the wall is:
F = ∆p / ∆t = 5 kg m/s / 0.1 s = 50 N

Hence, the force exerted on the wall by the ball is 50 Newtons.