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.

[Give answer to 2 s.f.]

b) Analyse the motion through the air to find where the ballbearing lands. What is your calculated value for x? (2 s.f.)

Explanation / Answer

a) Using energy conservation for two position,

initially spring PE + gravtitational PE + KE = Final spring PE + gravtitational PE + KE

200 (0.1^2) /2 + 0 + 0 = 0 + mgLsin20 + mv^2 / 2

1 = (0.050 x 9.8 x 0.30 x sin20) + (0.050 v^2 /2 )

v = 6.16 m/s

b) height of ball, h = 0.30sin20 = 0.103 m

in vertical,

initial vertical velocity, uy = 6.16sin20 = 2.11 m/s

acc, ay = - 9.8 m/s

y = - 0.103 m

Using , y = uy*t + ay*t^2 /2

- 0.103 = 2.11t - 9.8t^2 /2

4.9t^2 - 2.11t - 0.103 = 0

t = 0.475 s

deltaX = ux * t = (6.16cos20) x 0.475

     = 2.75 m

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