Posted by **Alphonse** on Thursday, November 22, 2012 at 11:59pm.

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?

- Physics -
**drwls**, Friday, November 23, 2012 at 9:34am
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

- Physics -
**Alphonse**, Friday, November 23, 2012 at 6:30pm
For some reason, it says it's the wrong answer.

- Physics -
**Mike**, Saturday, November 24, 2012 at 4:03pm
Hmmm well I've tried it and, for what it's worth, the way that drwls has done it is right.

- Physics -
**Alphonse**, Thursday, November 29, 2012 at 8:02pm
Yeah you're right, it does work. I must have done something wrong the first time.

Thanks!

- Physics -
**Jalopy **, Thursday, November 29, 2012 at 9:36pm
I don't get how you got the right answer...?

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