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Consider a snooker ball, with a mass of 0.07kg, rolling along a table at a speed of 4.1m/s. It hits a side cushion of the snooker table in a head-on elastic collision.

If the ball and cushion are in contact for 72 milliseconds, what is the size of the net force applied to the ball during this collision, in newtons?

Respuesta :

Answer: [tex] - 3.4 \times 10^{-3} N[/tex]

Explanation:

From Newton's law of motion, Force is equal to rate of change of momentum.

The initial velocity of the ball, u = 4.1 m/s

mass of the ball, m = 0.07 kg

During contact, velocity = 0

Time of contact, t = 72 ms = 72 × 10⁻³ s

[tex]F = \frac{m(v-u)}{t} = \frac{0.07 kg \times (0-4.1 m/s)}{72 \times 10^{-3}s} = - 3.4 \times10^{-3} N[/tex]

Thus, net force applied to the ball during this collision is [tex] - 3.4 \times 10^{-3} N[/tex]

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