Exam2 practice solution

# Exam2 practice solution - Mar06 LeClair PH 105-3 Exam 2...

This preview shows pages 1–4. Sign up to view the full content.

Mar06 LeClair PH 105-3 Exam 2 Solutions Instructions: There are 8 problems below. Choose any 6 problems to solve. All questions have equal point values. ± Problem 1: Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (see below). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. Find the original elastic potential energy in the spring if M = 0.250 kg. What we want in the end is to relate the initial energy (±gure part a) stored in the spring to the ±nal energy which is purely kinetic. Thus at some point conservation of total energy is needed. What we ±rst have to realize is that once the cord has been burned, the spring pushes away both masses, and the masses are moving apart in the ±nal state. This means that each block has kinetic energy after the cord is cut, not just the one with mass 3M. Therefore, we need to ±nd the velocity of the block of mass M before we proceed. Finding the velocities after the cord has been cut can be most easily done by conservation of momentum: 1

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

View Full Document
- * p before = - * p after - * p before = 0 - * p after = 3M v 3M + M v M v M = - 3 v 3M = - 6m / s After we have both velocities, we can use conservation of energy: K i + U i = K f + U f 0 + U spring = K f + 0 K f = 1 2 M v 2 M + 1 2 (3M) v 2 3M K f = (0 . 5)(0 . 25)( - 6) 2 J + (0 . 5)(3)(0 . 25)(2) 2 J K f = 4 . 5J + 1 . 5 J = 6J U spring = 6J ± Problem 2: A bead slides without friction around a loop-the-loop (below). The bead is released from a height h = 2 . 95 R . What is its speed at point A? Answer in terms of R and g , the acceleration of gravity. Since there is no friction present, we are only dealing with conservative forces, i.e., forces indepen- dent of the path taken. Basically, this means that we can use conservation of energy and make this a very simple calculation. Since the object starts out at rest at a height h oF of the ground, its initial energy is purely potential: E i = K i + U i = mgh When it reaches the top of the loop-the-loop, it has some velocity v and a height oF the ±oor of 2 R . The total energy is now: 2
E f = K f + U f = 1 2 mv 2 + mg (2 R ) The most common mistake here was to forget that at point A on the loop, the height oF of the ground is not zero and miss the 2 R . Equating E i and E f , mgh = 1 2 mv 2 + 2 mgR mg ( h - 2 R ) = 1 2 mv 2 g ( h - 2 R ) = g (2 . 95 R - 2 R ) = 0 . 95 gR = 1 2 v 2 v 2 = 2(0 . 95) gR v = p 1 . 9 gR Note that if we had measured potential energy relative to the top of the loop instead of the ±oor, we could just write down E i = mg ( h - 2 R ) and E f = 1 2 mv 2 and the result is the same. It has to be, gravitational potential energy only depends on height diferences .

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.

## This note was uploaded on 04/28/2008 for the course PHY 1102 taught by Professor Sawyer during the Spring '08 term at FIT.

### Page1 / 9

Exam2 practice solution - Mar06 LeClair PH 105-3 Exam 2...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online