Question

The body collides inelastically with the wall. In such a collision, with an initial speed v, the body's temperature increases by 0.5 K. By how much would the body's temperature increase with an initial speed of 4v, assuming that during a collision, half of the body's kinetic energy is always converted into the body's internal energy?

Answer 1

The body's temperature increases in an inelastic collision. We can use a formula to calculate the change in temperature.

Step-by-step explanation:

In an inelastic collision, the body's kinetic energy is not conserved. Instead, some of it is converted into internal energy, resulting in a temperature increase. Since half of the body's kinetic energy is always converted into internal energy, we can use the formula:

ΔT = 2v2 * C / m

where ΔT is the change in temperature, v is the initial velocity, C is the specific heat capacity, and m is the body's mass. Let's say the change in temperature for an initial velocity of v is 0.5 K. To find the change in temperature for an initial velocity of 4v, we substitute 4v into the formula:

ΔT' = 2(4v)2 * C / m

Learn more about Inelastic collisions