Question

A small steel ball bearing with a mass of 11.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.35 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 27.0^o from the horizontal.

Answer 1

Solution: The mass of the ball bearing m = 11.0g = (11.0/1000)kg = 0.011 kg.

Let k be the spring constant, the elastic potential energy is stored when the spring is compressed. It is given by

U = (1/2)kr² where r is the displacement of free end of spring.

Since small steel ball is shot from the compressed spring; spring’s elastic potential energy is converted into steel balls kinetic energy [K = (1/2)*mv²] and as the ball goes upward the kinetic energy is converted into ball gravitational potential energy U_g

U_g = mgh;

where h is the maximum height the steel ball reaches and h = 1.35m

From the above discussion it follows that,

Elastic potential energy = Kinetic energy of ball = Gravitational potential energy of ball

U = K = U_g

We can use K = U_g to calculate the initial launch speed of the ball

(1/2)*mv² = mgh

v² = 2gh

v = √[2gh]

v = √[2*(9.81m/s²)*(1.35m)]

v = 5.1466 m/s

Thus the compressed spring’s elastic potential energy is converted into the kinetic energy of the ball and its initial speed is v = 5.1466 m/s.

When the same spring is aimed at the angle θ = 27^o from horizontal, the steel ball has the same initial speed when the compressed spring is released. The horizontal distance (i.e. the range of projectile) covered by the by the steel ball is given by,

R = v²*sin(2θ)/g

R = (5.1466 m/s)²*sin(2*27^o) / (9.81m/s²)

R = 2.1844 m

Thus the ball would travel 2.18 m if the same spring were aimed at 27^o from the horizontal.