Question

A block 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

Answer 1

The correct answer is: 6.7 m/s  (recheck the options of the question)

Explanation:
For the perfectly elastic collision you can use the following shortcut:

$v_{1i} + v_{1f} = v_{2i} + v_{2f}$

9.2 - 2.5 = 0 + $v_{2f}$

Minus sign is used to indicate that the direction is reversed. The initial  velocity of the first block is 9.2 and its final velocity is 2.5 (with reversed direction); the initial velocity of the second block is 0.


$v_{2f}$ = 6.7 m/s 

Answer 2

Answer:

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)