A spring (k=4200N/m) and box A (m A =120kg) are on a frictionless incline, as sh

Question

A spring (k=4200N/m) and box A (mA=120kg) are on a frictionless incline, as shown below . Box A is pressed against the spring such that it is compressed 1.0m, and then released. Box A then hits, and sticks to box B, 1.0m farther up the ramp from the uncompressed position of the spring (massB=80kg) (i.e. the collision happens 2.0m total distance from A's initial position). Box B is held at rest until it is struck by A, then it is free to move. How far up the ramp will the two of them travel up the ramp before starting to slide back down?

Explanation / Answer

V = velocity of block A just before hitting block B

h = height gained by block A as it moves 2 m along the incline = 2 Sin30 = 1 m

Ma = mass of block A = 120 kg

using conservation of energy for block A

Elastic potential stored in spring = Potential energy + Kinetic energy of block A

(0.5) k x2 = mgh + (0.5) m V2

(0.5) (4200) (1)2 = 120 x 9.8 x 1 + (0.5) (120) V2

V = 3.92 m/s

Using conservation of momentum

V' = combined velocity of two block after collision

Ma V = (Ma + Mb) V'

120 x 3.92 = (120 + 80) V'

V' = 2.35 m/s

using conservation of momentum ,

H = height gained by the combined mass above height 'h"

Kinetic energy after collision = Potential energy at height gained by the combined mass above height 'h"

(0.5) (Ma + Mb) V'2 = (Ma + Mb) gH

(0.5) (2.35)2 = (9.8) H

H = 0.282 m

distance travelled up the ramp = H/Sin30 = 0.282 / 0.5 = 0.564 m

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