At collision use conservation of momentum
Mbullet Vi bullet = (Mbullet+MBlock)V
solve for V, the speed of the block with the bullet in it
calculate kinetic energy
(1/2) m v^2
that becomes potential energy when the block stops
m g h
so
v^2 = 2 g h
solve for h
now get angle
hypotenuse is L, length of string
T = string angle from vertical
cos T = (L-h)/L
solve for T
by the way you can approximate cosine for small angles
Cos T = 1 - T^2/2
1 - T^2/2 = 1 - h/L
T^2 = 2 h/L
T = sqrt(2 h/L)
note T will be in RADIANS if you do that
A bullet (m = 0.0250 kg) is fired with a speed of 92.00 m/s and hits a block (M = 2.50 kg) supported by two light strings as shown, stopping quickly. Find the height to which the block rises.
Find the angle (in degrees) through which the block rises, if the strings are 0.540 m in length.
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