Question

A bullet of mass
mb = 13 g
is fired with a speed of 360 m/s at a target that is a sheet of metal
(mass mt = 490 g).
The target is square with sides of length
L = 29 cm.
The target is hinged along its top edge so that it can swing freely (see figure below). If the bullet hits in the center of the target and then becomes stuck, what is the angular velocity of the target plus bullet just after the bullet hits?

Answer 1

So you would have the conservation of momentum in the first part, then when it starts to swing you could use the velocity (tangential) to find the angular velocity.

It's an inelastic collision, so: \(m_b*v_i=(m_b+m_t)v_f\)

Then \(v_f=v_t=\omega *r\rightarrow \omega=\dfrac{v_t}{r}\)

Answer 2

ok so i ended up getting vf =(.013*360)/(.013+.49) which equals to approximately 9.3
i understood how to get w but i'm not sure what value i should use for r

Answer 3

Not sure what the image might show that is not represented in the question.