A block of mass 10g slides down a curved incline of height h=1m. The incline is

Question

A block of mass 10g slides down a curved incline of height h=1m. The incline is fixed to the ground. At the bottom the 10 g mass collides elastically with a second block (M=50g), which is initially at rest. Calculate the height reached on the incline by the block "m" after collision. lgnore friction everywhere.

Explanation / Answer

when the 10g mass reaches gnd its pot energy mgh is converted into its kinetic energy mv2/2 => mv2/2 = mgh => v =v(2gh) = 4.47 m/s is the velocity attained by the block after reaching down. let initial vel of 10g be v0 (= 4.47 m/s) and final vel of 10g be v1 and that of 50g be v2 conservation of momentum => intial momentum = final momentum => 10 v0 + 0 = 10 v1 + 50 v2 and since it is an elastic collision v2 - v1 = v0 - 0 (http://en.wikipedia.org/wiki/Elastic_collision => check this out for the derivation) we got the two eqations so solving them gives v1 = - 2 v0 /3 = 2.98 m /s so the final velocity of the 10g bloc is 2.98 m/s which is used to climb up the incline therfore using v1 = v 2 gh1 => h1 = .435 m is the height to which the 10g reach the incline.

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