A ball of mass m is shot straight up into the air by a spring-loaded launcher. I

Question

A ball of mass m is shot straight up into the air by a spring-loaded launcher. Initially, the spring is compressed by a distance D. After the spring is released, the ball has a velocity v out of the launcher and finally reaches a maximum height H. Ignoring air resistance, which of the following statements are true?

True False  If the spring constant is doubled, the ball will max out at height 8H

True False  The initial potential energy of the spring is equal to the final gravitational potential energy of the ball

True False  If the compression of the spring is doubled, the velocity of the ball out of the launcher will be 8v

True False  If the compression of the spring is doubled, the ball will max out at height 4H

True False  If the mass of the ball is doubled, the ball will max out at height 1/2 H

Explanation / Answer

According to given problem,

1.) Ans: Flase,

Reaon:  it would reach a height of 2H, Using the conservation of energy,

mgh = 1/2kx2

if k is double then height will slo be doubled.

2.) Ans: True.

Reason:After the ball leaves the launcher, the sum of the ball's PE and the ball's gravitational PE is constant. Conservation of mechanical energy

3.) Ans: Flase,

Reason: If the compression of the spring is doubled, the velocity of the ball out of the launcher will be 2v
because 1/2k*y2 = 1/2*m*v2 so if y is doubled so is v.

4.) Ans: True.

Reason: If the compression of the spring is doubled, the ball will max at the height 4H,

Because 1/2ky2 = mgh so if y is doubled then height will go to 4H.

5.) Ans: True,

Reason: Using conservation of energy principal,

is the mass is doubled the max heigh becoms H/2.

I hope you understood the problem, If yes rate me!! or else comment for a better solution.  

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