Question

A body of mass 4 kg moving with a velocity 12 metre per second collides with another body of mass 6 kg at rest if the two bodies stick together after collision then the loss of an kinetic energy of a system is

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Mass $\sf{(m_1)}$ = 4kg

• Velocity $\sf{(u_1)}$ = 12m/s

• Mass $\sf{(m_2)}$=6kg

• Velocity $\sf{(u_2)}$ = 0

$\large\underline{\underline{\sf To\:Find:}}$

• Loss in kinetic energy = ?

$\large\underline{\underline{\sf Solution:}}$

$\large{\boxed{\sf m_1u_1+m_2u_2=m_1v_1+m_2v_2}}$

$\large\implies{\sf 4×12=(4+6)v}$

$\large\implies{\sf v=4.8m/s }$

$\large\implies{\sf Kinetic\:Energy\:Before\: Collision=\frac{1}{2}m_1u_1^2}$

$\large\implies{\sf \frac{1}{2}×4×(12)^2}$

$\large\implies{\sf 288J}$

$\implies{\sf Kinetic \:Energy \:After Collision =\frac{1}{2}×(m_1+m_2)v^2}$

$\large\implies{\sf 10×(4.8)^2 }$

$\large\implies{\sf 115.2J}$

Loss in Kinetic Energy = 288 - 115.2

$\implies{\sf 172.8J }$

$\large\underline{\underline{\sf Answer:}}$

•°• Loss in Kinetic Energy is 172.8J

Answer 2

Answer:

172.8J

Explanation:

Loss in kE energy =m1*m2(u1-u2)^2/2(m1+m2)

Where u1 and u2 are the intial velocity of object

In this question u1=12 u2=0