A .023 kg gold ball moving at 16.8 m/s crashes through the window of a house in 5.1x10^-4s.

A .023 kg gold ball moving at 16.8 m/s crashes through the window of a house in 5.1x10^-4s.

After the crash, the ball continues in the same direction with a speed of 10.2 m/s.

Assuming the force exerted on the ball by the window was constant, what was the magnitude of this force? Answer in units of N

First of all, we shall determine the acceleration, a, which is given by (v-u)/t, where v is the final speed, u is the initial speed and t is the reaction time between the ball and the window.

a= (10.2-16.8)/(5.1x10^-4)
= -1.294117647x10^4m/s^2

Resultant force exerted by the ball on the window, F
= ma, where m is the mass of the ball

F= (0.023)(-1.294117647x10^4)
= -297.6N

Thus, the magnitude of F= 300N correct to 2 sf.

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