A 2.4 kg block slides along a frictionless table at a speed of 1.5 m/s and colli

Question

A 2.4 kg block slides along a frictionless table at a speed of 1.5 m/s and collides elastically with a 0.24 kg block which is initially at rest. The surface of the table is 1 m above the floor.

(a) What is the speed of the 2.4 kg block after the collision?
(b) What is the speed of the 0.24 kg block (initially at rest) after the collision?
(c) Where how far from the base of the table does the 2.4 kg block land on the floor?
(d) Where how far from the base of the table does the 0.24 kg block land on the floor?

Explanation / Answer

1) v1 = (m1-m2)v1 /(m1+m2) = ( 2.4 - 0.24)*1.5/(2.4+0.24) = 1.23 m/s 2) v2 = 2m1v1/(m1+m2) = 2*2.4*1.5/(2.4+0.24) = 2.73 m/s 3) x = V1v(2h/g) = 1.23*v(2*1/9.81) = 0.5554 m = 55.54 cm 4) x = V2v(2h/g) = 2.73*v(2*1/9.81) = 1.23 m = 123.27 cm

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