Question

Topic: COLLISIONS. Impulsive forces

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a.) A stuntman of mass 77 kg belly-flops on a shallow pool of water from a height of 11 m. When he hits the pool, he comes to rest in about 0.050 s. What is the impulse that the water and the bottom of the pool deliver to his body during this time interval? What is the time-average force?

b.) A large ship of 7.0 x 10⁵ metric tons steaming at 20 km/h runs a ground on a reef, which brings it to a halt in 5.0 s. What is the impulse delivered to the ship? What is the average force on the ship? What is the average deceleration?

Answer 1

Part A.

Impulse is given by:

Impulse (J) = Change in momentum (dP)

Momentum = Mass*Velocity

So,

J = dP

J = Initial momentum - final momentum = Pi - Pf

J = m*Vi - m*Vf

Now we know that

Vf = final velocity of stuntman after he hits the pool = 0 m/sec ('he comes to rest')

m = mass of stuntman = 77 kg

for initial velocity of stuntman when he just hits the pool 'Vi', Using 3rd kinematic equation

Vi^2 = U^2 + 2*a*H

U = Velocity of stuntman when he jumps = 0

a = -g = -9.81 m/sec

H = height = -11 m

So,

Vi = sqrt (0^2 + 2*(-9.81)*(-11)) = 14.7 m/sec

So using this value:

Momentum = J = m*(Vi - Vf)

J = 77*(14.7 - 0) = 1131.9 kg-m/sec

Average force will be given by:

F_avg = Impulse/time of contact

time of contact = 0.050 sec

So,

F_avg = 1131.9/0.050

F_avg = 22638 N = 2.26*10⁴ N