Posted by Alphonse on .
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?
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)]
= 0.136 s
For some reason, it says it's the wrong answer.
Hmmm well I've tried it and, for what it's worth, the way that drwls has done it is right.
Yeah you're right, it does work. I must have done something wrong the first time.
I don't get how you got the right answer...?