A 2.4 kg block slides along a frictionless table at a speed of 1 m/s and collide

Question

A 2.4 kg block slides along a frictionless table at a speed of 1 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?
m/s

(b) What is the speed of the 0.24 kg block (initially at rest) after the collision?
m/s

(c) Where how far from the base of the table does the 2.4 kg block land on the floor?
m

(d) Where how far from the base of the table does the 0.24 kg block land on the floor?
m

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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