A ball of clay is moving at a speed of 12 m/s collides and sticks to a stationary ball of clay. If each ball has a mass of 13 kg, what's the combined velocity of the two balls of clay after the collision

p_{f} = 6 m / s

Explanation:

We can solve this exercise using conservation of momentum. For this we define a system formed by the two balls, so that the forces during the collision have been intense and the moment is preserved

Initial instant. Before the crash

        p₀ = m v +0

Final moment. Right after the crash

        = (m + m) v_{f}

how the moment is preserved

        p₀ = p_{f}

       m v = 2 m v_{f}

       v_{f} = v / 2

we calculate

       v_{f} = 12/2

       p_{f} = 6 m / s

the area between the graph and the t axis

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm