A 15kg block is attached to a very light horizontal spring of force constant 500

Question

A 15kg block is attached to a very light horizontal spring of force constant 500N/m and is resting on a frictionless horizontal table. Suddenly it is struck by a 3kg stone traveling horizontally at 8m/s to the right, whereupon the stone rebounds at 2m/s horizontally to the right. Find the maximum distance that the block will compress the spring after the collision.

Explanation / Answer

Initial momentum of the system : P initial = 8*3 (right) Final momentum of the system: let us assume that the stone imparts velocity of 'u' to the spring to the right. then, P final = 3*3 + 15*u Conservation of momentum, 8(3) = 3(3) + 15u => u = 1m/sec. Body is imparted velocity. Now, apply conservation of energy between spring and the body. Final stage, all the KE is converted into the PE of the spring. final velocity of the body is 0, at the maximum compression. hence, 1/2 * 15 * 1^2 = 1/2 * 500 * x^2 ( x is the max compression) => x= 0.1732m.

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