A ball with a mass of 120 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 15 (kg)/s^2 and was compressed by 3/5 m when the ball was released. How high will the ball go?

2 Answers
Jul 10, 2018

The height reached by the ball is =2.30m

Explanation:

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The spring constant is k=15kgs^-2

The compression of the spring is x=3/5m

The potential energy stored in the spring is

PE=1/2kx^2=1/2*15*(3/5)^2=2.7J

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

KE_(ball)=1/2m u^2

Let the height of the ball be =h

The acceleration due to gravity is g=9.8ms^-2

Then ,

The potential energy of the ball is PE_(ball)=mgh

Mass of the ball is m=0.120kg

PE_(ball)=2.7=0.120*9.8*h

h=2.7/(0.120*9.8)

=2.30m

The height reached by the ball is =2.30m

2.293\ m

Explanation:

Assuming spring to be perfectly elastic, when the ball is released, the entire elastic energy of spring 1/2kx^2 is transferred to the ball in form of kinetic energy 1/2mv^2
Where, k is spring constant,

x initial compression of spring

m is mass of ball

u is the velocity of ball when it is thrown vertical upward

hence, we have

\text{K.E. of ball}=\text{elastic energy of spring}

1/2(0.120)u^2=1/2(15)(3/5)^2

u^2=135/3

u=\sqrt{135/3}

Now, using third equation of motion:

v^2=u^2+2as

For the motion against gravity, we have a=-g=-9.81\ \text{m/s}^2, initial velocity u=\sqrt{135/3} & final velocity v at the maximum height s=h becomes zero i.e. v=0

hence, we have

0^2=(\sqrt{135/3})-2(9.81)h

h=\frac{135}{3\cdot 19.62}

=2.293\ m