- #1
FancyNut
- 113
- 0
I'm stuck on this problem..
My futile attempt:
the wood block's initial momentum is zero so it's [tex]m_b v_b = (m_b + m_w) v_f[/tex]
Where [tex]m_w[/tex] is the mass of the wooden block. In order to get the bullet's velocity (which is [tex]v_b[/tex]) I need [tex]v_f[/tex] which is the final velocity of both the bullet and the block in this inelastic collision.
So I tried to use kinematics to get that final velocity which is equal to initial velocity from the start of motion (bullet + block) until it comes to rest after moving .05 meters.
[tex]v_f^2 = v_i^2 + 2 x a[/tex]
[tex]0 = v_i^2 + 2 (.05) a[/tex]
As you can see, I don't know the acceleration... so maybe I should do some force analysis? The problem didn't give the magnitude of force the bullet exerted on the block nor the coefficients for friction so I don't know...
Thanks for any help!
A 10 g bullet is fired into a 10 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table.
What was the speed of the bullet?
My futile attempt:
the wood block's initial momentum is zero so it's [tex]m_b v_b = (m_b + m_w) v_f[/tex]
Where [tex]m_w[/tex] is the mass of the wooden block. In order to get the bullet's velocity (which is [tex]v_b[/tex]) I need [tex]v_f[/tex] which is the final velocity of both the bullet and the block in this inelastic collision.
So I tried to use kinematics to get that final velocity which is equal to initial velocity from the start of motion (bullet + block) until it comes to rest after moving .05 meters.
[tex]v_f^2 = v_i^2 + 2 x a[/tex]
[tex]0 = v_i^2 + 2 (.05) a[/tex]
As you can see, I don't know the acceleration... so maybe I should do some force analysis? The problem didn't give the magnitude of force the bullet exerted on the block nor the coefficients for friction so I don't know...
Thanks for any help!