Question

A 49.6 g marble moving at 3.99 m/s strikes a 13.6 g marble at rest.

A) What is the speed of the initially moving marble immediately after the collision?
2.2728 m/s

B)What is the speed of the initially not moving marble immediately after the collision?

Answer 1

Given initial condition

for particle 1 & for particle 2

m₁= 49.6 g , v_1i = 3.99m/s     m₂= 13.6 g , v_2i= 0m/s and let v_1f   & v_2f are final velocities.

Using momentum conservation law we get,

m₁v_1i +   m₂v_2i =  m₁v_1f   + m₂v_2f

=> 49.6x 3.99 + 13.6 X 0 = 49.6 v_1f x + 13.6 X v_2f

=> 197.90 = 49.6 v_1f x + 13.6 X v_2f ----------------------- 1

as given v_1f =2.2728m/s

then 197.90 = 49.6 x 2.2728 x + 13.6 X v_2f

v_2f  =6.26 m/s

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(ADDITIONAL INFO)

if v_1f =2.2728m/s was not given then

Using Kinetic energy conservation law we get,

(1/2) m₁v_1i ²+ (1/2)   m₂v_2i²=  (1/2) m₁v1f ²  + (1/2) m₂v_2f ²

=> 49.6 x 3.99² + 13.6 X 0 ² = 49.6 v1f ² x + 13.6 X v_2f 2

=> 789.63= 49.6 v1f ² x + 13.6 X v_2f ²----------------------- 2

solving the quadratic( by combining 1 & 2 will give the answer