Question

A 14.3 gram bullet embeds itself in a 2.823 kg block on a horizontal frictionless surface which subsequently collides with a spring of stiffness 378 N/m.  The maximum compression of the spring is 13.5 centimetres.
 
a)  What was the initial speed of the bullet in m/s?

b)  How long did it take for the bullet-block system to come to rest after contacting the spring?

Answer 1

a) Momentum is conserved during the embedding process. Kinetic energy is conserved during the spring compression by a distance X. Do the problem "backwards": first figure out the velocity after embedding but before comporession, using the elastic relationship for this process.

(1/2)*k*Xmax^2 = (1/2)(m+M)v^2
where v is the velocity after the bullet is embedded but before compression.
(378)*(0.135)^2 = 2.837*v^2
v = 1.558 m/s
Now let V be the initial bullet velocity. Apply conservation of momentum to the bullet-embedding process.
0.0143*V = (m+M)*v
0.0143*V = 2.837*1.558
V = 309 m/s

b)(1/4) of the period of the spring-mass system with the bullet added.

time to stop = (P/4)
= (1/4)(2 pi)*sqrt[(m+M/k)]
= (pi/2)*sqrt[2.837/378]
= 0.136 s

Answer 2

For some reason, it says it's the wrong answer.

Answer 3

Hmmm well I've tried it and, for what it's worth, the way that drwls has done it is right.

Answer 4

Yeah you're right, it does work. I must have done something wrong the first time.

Thanks!

Answer 5

I don't get how you got the right answer...?