The speed of the blocks after the collision is most nearly will be Vf= 2.376 m/s.
Collision is defined as when two bodies of different masses move towards each other with a particular speed, then there will be three cases after the collision.
Elastic collision, in which the kinetic energy of the object remains constant.
Inelastic collision in which the kinetic energy is not conserved after the collision.
Completely elastic collision, in which the masses stick together after the collision and move with a final speed.
It is given in the question that
Mass of body A= 2 kg
Velocity of body A= ?
Mass of body B=3 kg
Velocity of body B= 0 m/s
since the body is coming from the inclined place at an angle 30 so the height will be calculated as:
[tex]\rm Sin\theta=\dfrac{h}{H}[/tex]
Inclined plane H=3.6 m
[tex]\rm Sin30=\dfrac{h}{3.6}[/tex]
[tex]h=1.8 m[/tex]
Now the initial velocity of the object from the conservation of energy
[tex]m_Agh=\dfrac{1}{2}m_Av_A^2[/tex]
[tex]v_A=\sqrt{2gh[/tex]
[tex]v_A=\sqrt{2\times9.81\times 1.8}=5.94\ \frac{m}{s}[/tex]
Now form the collision is completely elastic so the momentum before collision will be equal to after collision.
[tex]m_Av_A+m_Bv_B=(m_A+m_B)v_f[/tex]
[tex](2\times 5.94)+(3\times0)=(2+3)v_f[/tex]
[tex]v_f=\dfrac{11.88}{5}=2.376\ \frac{m}{s}[/tex]
Thus the speed of the blocks after the collision is most nearly will be Vf= 2.376 m/s.
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