# Kinematics 1D

• I
Zerefss

## Main Question or Discussion Point

A bullet is ﬁred into a block of wood at a speed of 200 m/s. Upon hitting the block of wood the bullet decelerates with magnitude 640,000 m/s2 and the block accelerates with magnitude 360,000 m/s2. When the bullet and block obtain the same speed the accelerations stop. At this point, the bullet is inside the block, moving with the block.
(a) What is the speed of the bullet and block after the accelerations stop?

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I was browsing through older posts when I saw your question...
Let v denotes velocity of the bullet
v = v_o + a_o*t, v_o = 200 m/s, a_o = - 640,000m/s2
Let V denotes velocity of the wood block
V = V_o + A_o*t, V_o = 0 m/s, A_o = 360,000 m/s2
at some time t, v = V, we get
200 - 640,000t = 360,000t
1,000,000t = 200
t = 2x10^-4s
Substituting t into V equation, you should get V = 72m/s, which has same value as the speed of the bullet and block after the accelerations stop.

jim mcnamara
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