Question

Particle A has 300 J of kinetic energy. Particle B has 2 times the mass and 1/2 times the speed of particle A. What is the kinetic energy of particle B?

Answer 1

Given data

*The particle A has kinetic energy is U_a = 300 J

*Particle B has a mass is m_b = 2 × m_a

*Particle B has a speed is v_b = 1/2 × v_a

The formula for the kinetic energy of the particle B is given as

$U_b=\frac{1}{2}m_bv^2_b$

Substitute the known values in the above expression as

$\begin{gathered} U_b=\frac{1}{2}(2\times m_a)(\frac{1}{2}\times v_a)^2 \\ =\frac{1}{4}m_av^2_a \\ =\frac{1}{2}\times\frac{1}{2}m_av^2_a \\ =\frac{1}{2}\times300 \\ =150\text{ J} \end{gathered}$

Hence, the kinetic energy of the particle B is U_b = 150 J