A ball of clay of mass m = 0.32 kg is thrown in the x direction with an initial velocity of v = 19 m/s. It strikes a block of m² = 4.8 kg initially at rest and sticks. What is the final velocity of the clay and the block after the collision?



Answer :

Answer:

1.1875 m/s

Explanation:

Since no net external force act on the system, the total momentum of the system is conserved. Momentum is given by the equation [tex]p=mv[/tex] where [tex]m[/tex] is the mass of the body and [tex]v[/tex] is the velocity of the body. Since the clay sticks with the block after collision, the final velocities of the clay and the block are the same. Keeping all these in mind, let us form the equation for the conservation of momentum:

[tex]p_{final} =p_{initial}\\\\(p_{clay}+p_{block})_{final}=(p_{clay}+p_{block})_{initial}\\\\(m_{clay})(v)+(m_{block})(v)=(m_{clay})(u_{clay})+(m_{block})(u_{block})[/tex]

Substituting all the values from the question, we get:

[tex](m_{clay})(v)+(m_{block})(v)=(m_{clay})(u_{clay})+(m_{block})(u_{block})\\\\(0.32)(v)+(4.8)(v)=(0.32)(19)+(4.8)(0)\\\\5.12v=6.08\\\\v=1.1875 m/s[/tex]