Question

On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediately after the collision, the 0.200 kg block is moving at 12.0 m/s in the direction 30° north of east and the 0.400 kg block is moving at 13.0 m/s in the direction 53.1° south of east. Use coordinates where the x-axis is east and the y-axis is north. (a) What is the total kinetic energy of the two blocks after the collision (in joules)

Answer 1

Answer:48.2 Joules

Step-by-step explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

`v_(0.2)=12\cos (30)\hat{i}+12\sin (30)\hat{j}`

`|v_(0.2)|=11.99 m/s\approx 12 m/s`

Velocity of 0.4 kg mass

`v_(0.4)=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}`

`|v_(0.4)|=12.99 m/s\approx 13 m/s`

Thus total Kinetic energy
`=(0.2* 12^2)/(2)+(0.4* 13^2)/(2)`

Kinetic energy=48.2 J