Ablock of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s it collides in a perfectly elastic collision with a second block of mass m that is at rest. after the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s what is v, the speed of the second block after the collision? a)6.4 m/s b)5.1 m/s c)5.9 m/s d)7.2 m/s

option (A)

Explanation:

m = 4.4 kg, u = 9.2 m/s, v = - 2.5 m/s

M =

U = 0

V = ?

By use of conservation of momentum

mu + M x 0 = mv + MV

4.4 x 9.2 + 0 = 4.4 x (- 2.5) + M x V

40.48 + 11 = M V

MV = 51.48       (1)

By using the conservation of kinetic energy

0.5 x m x u^2 + 0 = 0.5 x m v^2 + 0.5 x M V^2

4.4 x 9.2 x 9.2 = 4.4 x 2.5 x 2.5 + MV^2

372.416 - 27.5 = MV^2

MV^2 = 344.9   (2)

Dividing equation (2) by equation (1)

V = 6.7 m/s

So, the correct option is (A)

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