Question

A 0.248 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.148 kg puck moving initially along the x axis with a speed of 2.82 m/s. After the collision, the 0.148 kg puck has a speed of 1.82 m/s at an angle of 25◦ to the positive x axis. Find the fraction of kinetic energy lost in the collision.

Answer 1

The fraction of kinetic energy lost in the collision is 0.25 .

Step-by-step explanation:

We know , momentum will be conserved along x direction .

Therefore ,

`0.248 * 0+0.148* 2.82 = 0.148* 1.82 * sin \ 25^(\circ) + 0.248* v\\\\0.42=0.11+0.248 * v\\\\v=1.25\ m/s`

Now , fraction lost in kinetic energy is :

`Loss =(K.E_i-K.E_f)/(K.E_i)\\\\Loss=1-(K.E_f)/(K.E_i)\\\\Loss=1-((0.148*1.82^2)/(2)+(0.248 * 1.25^2)/(2))/((0.148* 2.82^2)/(2)+(0.248* 0^2)/(2))\\\\Loss =0.25`

Therefore , the fraction of kinetic energy lost in the collision is 0.25 .