Perfect Inelastic collision with string tension

[erm151]

Homework Statement

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 0.0132 kg is fired at the block with a horizontal velocity v-_i. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.92 N . What is the v_i of the bullet?

Homework Equations

Tension=mg+ma
a=mv^2/r
(m_1)(v_1)=(m_1+m_2)v_2

The Attempt at a Solution

First I input the centripetal acceleration equation into the tension equation so T=mg+m(((m_1+m_2)(v_2)^2)/r) then plugged in the numbers given in the equation so the formula began to look like 4.92N=(0.8132)(-9.8)+0.8132((0.8132(v_2)^2)/1.6) and when I solved for v_2 I got approximately 5.584. Then I used the 5.584 in the conservation of motion equation so v_1=(0.8132*5.584)/0.0132. That gives me 344m/s for velocity of the bullet but mastering physics says that wrong.

 

[BvU]

Hello erm,

Good start ! Good use of the template. Clear post. Pleasure to assist.

You want to take into account that the tension in the cord has a different direction than the $mg$: it's all vectors.

Furthermore, your $v_2$ is at the bottom; if the block swings up, the speed will decrease. Exercise doesn't give that it hangs still after rising 0.8 m -- that would be too easy I suppose.

This is a nice, but not so easy exercise. You'll have to work it out in a few steps. Momentum balance for the initial inelastic collision is a good start. What do you want to get out of it (in the form of an expression with symbols) ?

 

[erm151]

So for the tension of the cord part would I have to include the cos(theta) into the work so T=mg(cos(theta)+ma? I don't understand what you mean by what I want to get out of the initial inelastic collision? If I understand you correctly I want to get the velocity of the bullet from the equality so v_1=((m_1+m_2)v_2)/m_1

 

[BvU]

OK, so you have a relationship between $v_2$ and the speed of the bullet. That's part of what I meant. For the next step you have to relate this $v_2$ to something else.

For the other part, I think you'll make things much clearer for yourself with a free-body diagram of the block at the point when it's risen 0.8 m.

 

[erm151]

Ok I figured it out. I had to find out the velocity at the height of 0.8m then use the KE+PE=Total energy equation so that 1/2 mv^2+mgh=1/2mv^2+mgh and one side is the velocity at 0.8 and the other (the velocity I am looking for) is calculated at h=0. so 1/2(0.8132)(v^2)=1/2(0.8132)(v at 0.8)+0.8132(9.8)(0.8)