A 0.50 kg block travels at 2.00 m s⁻¹ on a frictionless horizontal surface. It collides with a 1.00 kg block, and after impact the two stick together and move as one. The amount of kinetic energy dissipated in the collision is

Question

Accepted Answer

0.67 J. The kinetic energy before the collision is calculated using the formula KE = 0.5 * m * v² for the moving block, and after the collision, the combined mass moves with a new velocity. The difference in kinetic energy before and after the collision gives the amount of energy dissipated, which in this case is 0.67 J.

A 0.50 kg block travels at 2.00 m s⁻¹ on a frictionless horizontal surface. It collides with a 1.00 kg block, and after impact the two stick together and move as one. The amount of kinetic energy dissipated in the collision is

Solution

Related JEE Main Physics questions