Question

A spring-loaded toy dart gun is shot to a height h. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed twice as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring

Answer 1

Answer:

The height reached by the dart in the second shot is (4 H).

Explanation:

It is given that, a spring-loaded toy dart gun is shot to a height h. In this case, all the potential energy stored in the spring is converted to potential gravitational energy at the maximum height.

$\dfrac{1}{2}kx^2=mgH$........(1)

At the second shot, the spring is compressed twice as far before firing. x' = 2x

$\dfrac{1}{2}kx'^2=mgh$

$\dfrac{1}{2}k(2x)^2=mgh$.........(2)

h is the height reached by the dart in the second shot.

Dividing equation (1) and (2) as:

$4=\dfrac{h}{H}$

h = 4H

So, the height reached by the dart in the second shot is (4 H). Hence, this is the required solution.

Answer 2

Answer:

Explanation:

Let the spring constant is K and the mass of the shot is m.

Case 1: when the compression in the spring is d.

Use the energy conservation

Elastic Potential energy of the spring = gravitational potential energy of the shot

1/2 K x d² = mgh   .... (1)

Case 2: Let the shot rises upto height h'.

1/2 K x (2d)² = mgh' .... (2)

Divide equation (2) by equation (1)

h' / h = 4

h' = 4h

Thus, the shot rises by the four times the initial height.