Question

In order to sell mattresses, companies will often drop a bowling ball on a mattress. Let's assume a 5kg bowling ball is dropped 1.2m above the bed.

A. What is the speed of the bowling ballright before it hits the bed? Ignoring air resistance.

B. The ball hits the bed (approximated as a spring, assume it follows Hooke's law) and depresses the bed a maximum of .127m (about half an inch). What is the spring constant of the bed?

C. After crashing in to the bed, the ball is propelled back to a max height of .31m. How muchwork was done by non-conservative forces on the ball?

Answer 1

by third law of motion

v^2 - u^2 = 2*g*s where v = final velocity, u = initial velocity, s = displacement, g = 9.8 m/s^2

v^2 - 0 = 2*9.8*1.2

v = 4.85 m/s Velocity before hitting the bed

part B)

K.E. of ball = Energy transferred to spring

0.5*m*v^2 = 0.5*k*x^2

5*4.85*4.85/(0.127*0.127) = k = 7291.989 N/m ~ 7292 N/m(Rounded off)

Part c)

Work done = Force*displacement = mass*acceleration*displacement

by third law of motion

v^2 - u^2 = 2*a*s

0 - 4.85*4.85 = 2*(g-a)*0.31

g - a = -37.94 m/s^2

a = 9.8+37.94 = 47.74 m/s^2

Work done = 5*47.74*0.31 = 73.997 J