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.

a. What is the average force on the first ball? Express your answer with the appropriate units. Enter positive value if the force is along the first ball's initial speed and negative value if the force is opposite to the first ball's initial speed.

b. What is the average force on the second ball? Express your answer with the appropriate units. Enter positive value if the force is along the first ball's initial speed and negative value if the force is opposite to the first ball's initial speed.

Answer 1

Since there is no external force applied, So momentum will remain conserved, So

Pi = Pf

m1*u1 + m2*u2 = m1*v1 + m2*v2

m1 = m2 = m, So

m*u1 + m*u2 = m*v1 + m*v2

u1 + u2 = v1 + v2

u1 = Initial speed of ball m1 = 7.2 m/s

v1 = final speed of ball m1 = 0 m/s

u2 = Initial speed of ball m2 = 0 m/s

So,

v2 = final speed of ball m2 = u1 + u2 - v1

v2 = 7.2 + 0 - 0 = 7.2 m/s

Now Average force on the 1st ball will be:

F_avg = dP/dt = m*dV/dt

for 1st ball

F_avg = m1*(v1 - u1)/dt

dt = collision time = 250*10^-6 sec

m1 = mass of billiard ball = 0.28 kg

So,

F_avg = 0.28*(0 - 7.2)/(250*10^-6)

F_avg = -8064 N = -8.1*10^3 N = Force applied on 1st ball

(If you need answer in two significant figure than use F_avg = -8.1*10^3 N)

for 2nd ball

F_avg = m2*(v2 - u2)/dt

dt = collision time = 250*10^-6 sec

m2 = mass of billiard ball = 0.28 kg

So,

F_avg = 0.28*(7.2 - 0)/(250*10^-6)

F_avg = 8064 N = 8.1*10^3 N = Force applied on 2nd ball

(If you need answer in two significant figure than use F_avg = 8.1*10^3 N)

Let me know if you've any query.