Question

A .420-kg block of wood hangs from the ceiling by a string, and a 0.0750-kg wad of putty is thrown straight upward, striking the bottom of the block witha speed of 5.74 m/s. The wad of putty sticks to the block. Is the mechanical energy of this system conserved? How high does the putty-block system rise above the original position of the block? Please solve using an energy approach.

Answer 1

In this case, you should use conservation of momentum

m1v1 = m2v2

Initially, the only momentum is in the putty flying upward

m1v1 = (0.075 kg)(5.74 m/s) = 0.430

After the collision, the block and the putty move together

m2v2 = (0.074 + 0.420 kg)*v2

So, applying the first equation

v2 = (0.074 kg)(5.74 m/s) / (0.074 + 0.420 kg) = 0.871 m/s

Now that you have the initial velocity, you can solve for the distance it rises using

v_f^2 = v_0^2 + 2a*y

v_f, the final velocity, is zero, so

y = - (0.871 m/s)^2 / (2 * - 9.8 m/s/s) = 0.0387 m = 3.87 cm

Hope this helps!