Question

Object A has 25 J of kinetic energy. Object B has one-quarter the mass of object A.By what factor does the speed of each object change if total work -20 J is done on each?

Answer 1

Answer:

Explanation:

Kinetic energy of A, K = 25 J

let the mass of A is m, the mass of mass B is m/4.

Let the velocity of A is v.

The change in kinetic energy is same for both as the work done is same.

$\frac{1}{2}m_{A}v_{A}^{2}=\frac{1}{2}m_{B}v_{B}^{2}$

$\frac{1}{2}mv^{2}=\frac{1}{2}\frac{m}{4}v_{B}^{2}$

So, vB = 2 v

Thus, the velocity of B is 2 times the velocity of A.

Answer 2

Answer:

The speed of object B is 2 times of speed of object A.

Explanation:

Given that,

Kinetic energy of object A = 25 J

Mass of object B $m_{B}=\dfrac{1}{4}m_{A}$

Work done = -20 J

We need to calculate the factor of the speed of the object

Suppose the final kinetic energy is same for both object.

$K.E_{f_{A}}=K.E_{f_{B}}$

$\dfrac{1}{2}m_{A}v_{A}^2=\dfrac{1}{2}m_{B}v_{B}^2$

Put the value into the formula

$\dfrac{1}{2}m_{A}\times v_{A}^2=\dfrac{1}{2}\times\dfrac{1}{4}m_{A}\times v_{B}^2$

$\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}$

$v_{A}=\dfrac{1}{2}v_{B}$

$v_{B}=2v_{A}$

Hence, The speed of object B is 2 times of speed of object A.