Question

Mickey, a daredevil mouse of mass 0.0215 kg , is attempting to become the world's first "mouse cannonball." He is loaded into a spring-powered gun pointing up at some angle and is shot into the air. The gun's spring has a force constant of 60.9 N/m and is initially compressed a distance of 0.137 m from its relaxed position. If Mickey has a constant horizontal speed of 2.35 m/s while he is flying through the air, how high above his initial location in the gun does Mickey soar? Assume ?=9.81 m/s2 .

Answer 1

Let the mass of the mickey is m = 0.0215 kg

Let the spring constant be k = 60.9 N/m

Compression of the spring is x = 0.137 m

Initial horizontal velocity of the mouse is V_x = 2.35 m/s

Initial vertical velocity be V_y

When the spring is compressed , work done on it is stored in it as elastic potential energy

Work done is given by W = 1/2 k x²

= 1/2 * 60.9 * 0.137 * 0.137

= 0.5715 J

This work done or potential energy is equal to initial kinetic energy of the mouse

If V_o is the initial velocity of the mouse , then we can write ,

Elastic potential energy = Initial kinetic energy

1/2 * m V_o² = W

V_o² = 2 W / m

= 2 * 0.5715 / 0.0215

= 53.164

So initial velocity V_o = 53.164 = 7.29 m/s

In terms of horizontal and vertical components , initial velocity can be written as

V_o² = V_x² + V_y²

V_y² =  V_o² - V_x²

V_y = ( 7.29² - 2.35² )

=  ( 53.164 - 5.5225 )

=  47.641

= 6.90 m/s

This would be the initial vertical velocity of the mouse

By the time , mouse reaches the maximum height it's initial vertical kinetic energy is changed into potential energy

If h is the maximum height reached by the mouse , then we can write ,

1/ 2 * m V_y² = m* g * h

h = V_y² / 2 g

= 6.90² / ( 2 * 9.81 )

= 47.641 / 19.62

= 2.428 m

So the mickey soar at a height h = 2.43 m