Homework 12 - Solutions

# Homework 12 - Solutions - homework 12 – FRENNEA KYLE –...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 12 – FRENNEA, KYLE – Due: Apr 18 2007, 4:00 am 1 Question 1 part 1 of 3 10 points A block of mass 0 . 13 kg is attached to a spring of spring constant 23 N / m on a fric- tionless track. The block moves in simple har- monic motion with amplitude 0 . 18 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately be- fore it strikes the block is 57 m / s and the mass of the bullet is 2 . 6 g. 23 N / m . 13 kg 2 . 6 g 57 m / s Find the speed of the block immediately before the collision. Correct answer: 2 . 39422 m / s (tolerance ± 1 %). Explanation: Let : M = 0 . 13 kg , m = 2 . 6 g , A = 0 . 18 m , and k = 23 N / m . Since the block moves in simple harmonic motion, we know that x = A sin( ω t + ϕ ) and since the only force acting on it is the spring force F s =- k x , the equation of mo- tion F s = M a is M a = M d 2 x dt 2 =- k x , so the angular velocity of this initial motion is ω = radicalbigg k M = 13 . 3012 rad / s . The velocity at a given instant is found by differentiation v = d x dt = A ω cos( ω t + ϕ ) . The equilibrium position has the highest ve- locity, and the magnitude of the cosine func- tion is never greater than 1.0, so v eq = A ω = 2 . 39422 m / s . Question 2 part 2 of 3 10 points If the simple harmonic motion after the collision is described by x = B sin( ω t + φ ), what is the new amplitude B ? Correct answer: 0 . 263088 m (tolerance ± 1 %). Explanation: Convert the mass of the bullet to kg. Then use conservation of momentum M v eq + mv = ( M + m ) v f , since the bullet stays inside the block after collision. Both terms on the left have the same sign since the block was going from left to right. Solving for v f , we obtain v f = M v eq + mv M + m = (0 . 13 kg) (2 . 39422 m / s) (0 . 13 kg) + (0 . 0026 kg) + (0 . 0026 kg) (57 m / s) (0 . 13 kg) + (0 . 0026 kg) = 3 . 46493 m / s . The system is in harmonic oscillation after the collision as well, but with a different frequency and amplitude. The new frequency is, using the new (total) mass, is ω f = radicalbigg k M + m = radicalBigg (23 N / m) (0 . 13 kg) + (0 . 0026 kg) = 13 . 1702 rad / s , homework 12 – FRENNEA, KYLE – Due: Apr 18 2007, 4:00 am 2 and the velocity is now v = B ω f at equilib- rium. But this velocity is just the velocity immediately after the collision, since it oc- curred at equilibrium, so v f = B ω f , solving for B B = radicalbigg M + m k v f = radicalBigg (0 . 13 kg) + (0 . 0026 kg) (23 N / m) × (3 . 46493 m / s) = 0 . 263088 m . Alternative Solution: An equivalent ap- proach (once we have v f ) is conservation of energy, using the fact that the potential en- ergy of the spring once it is fully compressed (maximum amplitude) is E p = (1 / 2) k B 2 , or 1 2 ( M + m ) v 2 f = 1 2 k B 2 , so by solving for B we arrive at the same result....
View Full Document

{[ snackBarMessage ]}

### Page1 / 14

Homework 12 - Solutions - homework 12 – FRENNEA KYLE –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online