3. An 80 kg circus clown is shot from a cannon tilted at 45° above the horizonta
ID: 1411538 • Letter: 3
Question
3. An 80 kg circus clown is shot from a cannon tilted at 45° above the horizontal. He travels frictionlessly through the air and lands on some thick, air-filled mattresses lying on the ground 30 m from the cannon.
a) Use what you already know about projectile motion to find the clown’s speed as he leaves the end of the cannon.
b) Calculate the kinetic energy of the clown as he leaves the cannon.
c) If the clown was propelled by a spring compressed 0.9 m to shoot him out of the cannon and he moved frictionlessly in the cannon, use conservation of energy to calculate the spring constant of the spring used?
d) The potential energy of the spring when compressed is (1) greater than, (2) equal to, or (3) less than the kinetic energy of the clown as he leaves the cannon.
e) Calculate the kinetic energy of the clown at his maximum height and use this result and conservation of energy to calculate the maximum height reached by the clown.
f) Calculate the kinetic energy and the speed of the clown when he is at half of his maximum height.
j) What is the kinetic energy of the clown when he hits the mattresses lying on the ground? Use this kinetic energy and your knowledge that the mattresses were compressed 1.3 m to calculate the spring constant of the mattresses.
Explanation / Answer
A) Range of a projectile, R = v^2 sin 2theta /g
30 = v^2 sin90 degree /g
v = sqrt(30*9.8) = 17.146 m/s
B) KE =0.5 mv^2 =0.5*80*17.146^2 = 11760 J
C) Spring energy = KE + PE
0.5 kx^2 = 11760 + 80*9.8*0.9* cos 45 degree
0.5*k*0.9*0.9= 11760 + 499
k = (11760+499) /[ 0.5*0.9*0.9] = 30269 N/m
d) greater than because spring energy = KE + gravitational PE gained
e) horizontal velcity v' = 17.146 cos 45 degree
= 12.124m/s
KE at maximum height = 0.5mv'^2
=0.5*80*12.124^2
= 5880 J
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