A 45.0 gram bullet hits an initially stationary block of mass 3.955 kg which is

Question

A 45.0 gram bullet hits an initially stationary block of mass 3.955 kg which is resting against a spring at equilibrium (see figure below). The bullet is embedded in the block, and the pair have a speed of 4.50 m/s immediately after impact. (a) What is the initial energy of the block-bullet system (in Joules)? (b) Assuming that the surface on which the block is resting is frictionless, determine the spring constant if the block and bullet stop and reverse directions after travelling for 25.00 cm together. (c) How much time will the block need between when it is impacted by the bullet and when it returns to its initial position? (d) Suppose now that the surface has a coefficient of friction of 0.0297. By how much is the spring now compressed? (e) If the block is not physically connected to the spring, how far will it slide before coming to a stop? Note, this is the total distance travelled-not the final displacement from equilibrium-of the block. Answer in box. Show work in blank space on next page. Failure to do so results in a deduction of points. No exceptions. Answer to a) Answer to b) Answer to c) Answer to d) Answer to e) An additional page is attached on which you may show your work. Place all answers in the box provided, but show your work in the blank space outside of the box. Failure to do so results in a deduction of points. No exceptions.

Explanation / Answer

Energy of Block + Bullet system = 1/2 (M1 + M2) V2
M1 and m2 are masses of bullet and block respectively and V is their common speed
Hence Initial value of energy Eo = 1/2 (4)(4.5)2 = 40.5 J

B) When Block + bullet stop, their initial KE , Eo is converted into spring potential energy
Hence E0 = 1/2 k Xm2
40.5 = 1/2 k (0.25)2
k = 1296 N/m

c) Time between impact on block from bullet and when it returns to initial ( equilibrium position), is half of the time period of oscillation.
Time period of oscillation of spring mass system T = 2 pi (m/k)1/2
Hence required time t = T/2 = pi (m/k)1/2 = 3.14 * ( 4/1296)1/2 = 0.17 sec

d) As friction does -ve work on the system,
Initial KE + work done by friction = final potential energy
40.5 - 0.0297 *4*9.8 Xm = 1/2 * 1296 Xm2   ( work done by friction is - f mg Xm )
Xm = 24.91 cm

e) As block leaves the spring, spring is back to relaxed condition, that is there is no potential energy stored in spring. So
Work done by friction = change in KE = - Eo
- 0.0297 * 4 * 9.8 X = - 40.5 ( X is total distance travelled by block)
X = 34.78 m

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