Question

A billiard ball of mass 0.28 kg hits a second, identical ball at a speed of 7.2 m/s and comes to rest as the second ball flies off. The collision takes 250 μs. What is the average force on the first ball? The second ball?

Answer 1

Answer:

(a) -8064 N

(b) 8064 N

Explanation:

(a)

From Newton’s law of motion, Force, F=ma where m is mass and a is acceleration.

Since acceleration is the rate of change of velocity per unit time, then where v is velocity and the subscripts f and I denote final and initial

For the first ball, the mass is 0.28 Kg, final velocity is zero since it finally comes to rest, t is 0.00025 s and initial velocity is given as 7.2 s. Substituting these values we obtain

$F=0.28\times \frac {0-7.2}{0.00025}=-8064 N$

(b)

For the second ball, the mass is also 0.28 Kg but its initial velocity is taken as zero, the final velocity of the second ball will be equal to the initial velocity of the second ball, that is 7.2 m/s and the time is also same, 0.00025 s. By substitution

$F=0.28\times \frac {7.2-0}{0.00025}=8064 N$

Here, we prove that action and reaction are equal and opposite

Answer 2

The average force on the first ball is -8,064 N

The average force on the first ball is 8,064 N

The formula for calculating the average force according to Newton's second law is expressed as:

$F=ma\\F=m(\frac{v-u}{t} )$

m is the mass of the ball

v is the final velocity

u is the initial velocity

t is the time taken

Given the following parameters

m = 0.28kg

v = 0m/s

u = 7.2m/s

t = 250×10⁻⁶secs

Substitute the given parameters into the formula:

$F=0.28(\frac{0-7.2}{0.00025} )\\F=0.28(\frac{-7.2}{0.00025} )\\F=0.28(-28,800)\\F=-8,064N$

Hence the average force on the first ball is -8,064 N

For the average force on the second ball:

$F=0.28(\frac{7.2-0}{0.00025} )\\F=0.28(\frac{7.2}{0.00025} )\\F=0.28(28,800)\\F=8,064N$

Hence the average force on the first ball is 8,064 N

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