Physic energy question

Question

1.A bullet has 768 J of energy when traveling 201 m/s. What will the velocity of the bullet be when it has 185 J?

*answer and explanation please*

Answer 1

Kinetic Energy (E_K) = 1/2mv² let K = 1/2m then E_K = Kv²

If E_K = Kv² then K = E_K/v² and in this case K = 768/201² = 0.01900943

So, we can use that K for the same bullet at a different velocity. E_K = Kv² means v = √(E_K/K) = √(185/0.01900943) = √9732.0117 = 98.7 m/s

Answer 2

Kinetic Energy (E_K) = 1/2mv² let K = 1/2m then E_K = Kv²

If E_K = Kv² then K = E_K/v² and in this case K = 768/201² = 0.01900943

So, we can use that K for the same bullet at a different velocity. E_K = Kv² means v = √(E_K/K) = √(185/0.01900943) = √9732.0117 = 98.7 m/s

Answer 3

The Kinetic Energy of an object is equal to one half the mass multiplied by the Velocity squared, or:

KE = 1/2 mv²

The given information in the problem includes the KE, and the velocity.

Substitute those into the equation above and solve algebraically for the mass:

m = 2KE/v² = 0.038 kg (which is a reasonable mass for a bullet)

Then use the mass in the above equation along with the new K.E. of 185 Joules to solve algebraically for the velocity:

v = SQRT(2KE/m) = 98.7 m/s

This answer makes sense because since KE is proportional to the square of the velocity we would expect the KE to go down about 4 times (from 768 to 185 Joules) when the velocity goes down about 2 times (from 201 to 98.7 m/s)