Question

Two particles of mass 0.5 KG and 0.25 KG moving with velocity 4 m/s and -3 m/s collides head on in a perfectly inelastic collision find the velocity of the composite particle after the collision and the kinetic energy lost in the collision

Answer 1

Answer:

Velocity of composite particle = 1.67m/s

 Kinetic energy lost in collision = 4.1 J

Explanation:

Given: $m_{1}=0.5kg\\m_{2}=0.25kg\\v_{1}=4.0m/s\\v_{2}=-3.0m/s$

According to Principle of Conservation of Linear Momentum, v is the velocity of composite particle;

Formula: $(m_{1}+m_{2})v=m_{1}v_{1}+m_{2}v_{2}\\(0.5+0.25)v=(0.5*4)+(0.25(-3.0))\\0.75v=2+(-0.75)\\v=\frac{2-.75}{.75} \\=\frac{1.25}{.75} \\=1.67m/s$

Kinetic Energy lost in collision can be calculated as:

Formula: $K.E_{lost}=\frac{1}{2}(m_{1}v_{1}^{2})+\frac{1}{2}(m_{2}v_{2}^{2})-\frac{1}{2}(m_{1}+m_{2} )v^{2}$

$=\frac{1}{2}(0.5*4^2)+\frac{1}{2}(0.25(-3.0^2))-\frac{1}{2}(0.5+0.2)(\frac{5}{3})^2\\ =4+\frac{9}{8}-\frac{25}{24} \\=4.1 J$

Hope u find it easy now :)

Answer 2

Answer:

The velocity of the composite particle after the collision is 1.67 m/s and the kinetic energy lost in the collision is 4.083J

Explanation:

m1=0.5kg

m2=0.25kg

u1=4 m/s

u2= - 3.0 m/s  

According to law of conservation of momentum

m1u1+m2u2=(m1+m2)v

(m1+m2)v= 0.5 x 4 + 0.25 (-3.0)

v=$\frac{ 2 - .75}{0.75}$  =$\frac{1.25}{0.75}$

=5/3= 1.67 m/s

k.E final  =$\frac{1}{2}$ (m1+m2)v²

=$\frac{1}{2}$ x 0.75 x (5/3)²

=1/2 x( 3/4) (25/9)

=25/24

[K.E]initial = 1/2x 0.5 x 4²+1/2 x 0.25x(-3)²

=4+9/8

=41/8 J

Loss in K.E=K.E initial- K.Efinal

=41/8 -25/24

=49/12 J

=4.083J