2.6 kg block slides along a frictionless table at a speed of 1.2 m/s and collide

Question

2.6 kg block slides along a frictionless table at a speed of 1.2 m/s and collides elastically with a 0.26 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 2.6 kg block after the collision?
m/s

(b) What is the speed of the 0.26 kg block (initially at rest) after the collision?
m/s

(c) Where how far from the base of the table does the 2.6 kg block land on the floor?
m

(d) Where how far from the base of the table does the 0.26 kg block land on the floor?

Explanation / Answer

1) v1 = (m1-m2)v1 /(m1+m2)
= ( 2.6 - 0.26)*1.2/(2.6+0.26) = 0.98 m/s
2) v2 = 2m1v1/(m1+m2) = 2*2.6*1.2/(2.6+0.26) = 2.18 m/s

3) x = V1(2h/g)

       = 0.98*(2*1/9.81) = 0.4433 m = 44.33 cm

4) 

x = V2(2h/g)

       = 2.18*(2*1/9.81) = 0.9843m = 98.43 cm

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