A 50-g ballbearing is pushed up against a compressed spring (of spring constant

Question

A 50-g ballbearing is pushed up against a compressed spring (of spring constant k = 200 N/m) on a level piece of ground. When the ballbearing is released it rolls (leaving the spring behind, they were not attached), without slipping, up a 30-cm ramp at 20 degrees and flies off the top of the ramp.

a) How far from the base of the ramp does the ballbearing land (x), if the spring is initially compressed by s = 10 cm?

Use energy methods to work out how fast the ball is travelling as it leaves the top of the ramp.

The initial state can be spring compressed, ball stationary. The final state can be ball leaving top of ramp. You don't need anything in between.

Don't forget that the (spherical) ball has rotational kinetic energy as well as translational kinetic energy.

b) Analyse the motion through the air to find where the ballbearing lands. What is your calculated value for x?

The speed from the first part is the launch speed, but the launch is at 20 degrees, so you need to break the velocity into components.

If you neglect air resistance, then you can analyse the flight as a simple projectile motion.

Explanation / Answer

a)
use conservation of energy between bottom and top
spring potential energy at bottom = kinetic energy at top + gravitaional potential energy
0.5*K*x^2 = 0.5*m*v^2 + m*g*h
0.5*200*(0.1)^2 = 0.5*0.05*v^2 + 0.05*9.8* (0.3*sin 20)
1= 0.5*0.05*v^2 + 0.0503
0.5*0.05*v^2 = 0.9497
v = 6.16 m/s
Answer: 6.16 m/s

b)
This is at an angle of 20 degree
In vertical direction:
vi = 6.16*sin 20 = 2.11 m/s
a = -9.8 m/s^2
h = - 0.3*sin 20 m = -0.103 m
use:
h = vi*t + 0.5*a*t^2
-0.103 = 2.11*t - 0.5*9.8*t^2
4.9*t^2 -2.11*t - 0.103 = 0

solving or t, we get positive value of t = 0.475 s

for horizontal direction:
vi = 6.16*cos 20 m/s
t = 0.475 s
delta x = vi*t
= 6.16*cos 20 * 0.475
= 2.75 m
Answer: 2.75 m

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