{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# phyexam3 - Version 061 Midterm 3 orin(58140 This print-out...

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

Version 061 – Midterm 3 – florin – (58140) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A triangular wedge 5 m high, 14 m base length, and with a 11 kg mass is placed on a frictionless table. A small block with a 5 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below. 11 kg 14 m 5 m 5 kg Δ X wedge 11 kg 14 m 5 m 5 kg All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far Δ X wedge does the wedge slide to the right? 1. 5.2 2. 5.29412 3. 5.4 4. 5.85 5. 4.8 6. 5.71429 7. 3.82353 8. 4.375 9. 4.57143 10. 5.6 Correct answer: 4 . 375 m. Explanation: Let : M = 11 kg , m = 5 kg , L = 14 m , and H = 5 m . Consider the wedge and the block as a two- body system. The external forces acting on this system — the weight of the wedge, the weight of the block and the normal force from the table — are all vertical, hence the net hor- izontal momentum of the system is conserved, P wedge x + P block x = constant . Furthermore, we start from rest = center- of-mass is not moving, and therefore the X coordinate of the center-of-mass will remain constant while the wedge slides to the right and the block slides down and to the left, X cm = m X block + M X wedge m + M = constant . Note: Only the X coordinate of the center- of-mass is a constant of motion; i.e. , the Y cm accelerates downward because the P y compo- nent of the net momentum is not conserved. Constant X cm means Δ X cm = 0 and there- fore m Δ X block + M Δ X wedge = 0 . Note that this formula does not depend on where the wedge has its own center-of-mass; as long as the wedge is rigid, its overall dis- placement Δ X wedge is all we need to know. Finally, consider the geometry of the prob- lem: By the time the block slides all the way down, its displacement relative to the wedge is equal to the wedge length L , or rather - L because the block moves to the left of the wedge. In terms of displacements relative to the inertial frame of the table, this means Δ X block - Δ X wedge = - L . Consequently, 0 = m Δ X block + M Δ X wedge = m ( - L + Δ X wedge ) + M Δ X wedge and therefore Δ X wedge = m m + M L = (5 kg) (5 kg) + (11 kg) (14 m) = 4 . 375 m .

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

View Full Document
Version 061 – Midterm 3 – florin – (58140) 2 002 10.0 points Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 3 . 3 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s final speed is 1 . 1 m / s . 3 . 3 m / s 1 . 1 m / s θ φ Before After Find the eight ball’s angle φ with respect to the original line of motion of the queue ball. 1. 20.1713 2. 33.367 3. 30.7723 4. 34.8499 5. 26.3878 6. 23.1998 7. 18.8765 8. 35.2344 9. 16.3827 10. 19.4712 Correct answer: 19 . 4712 . Explanation: Let : v q i = 3 . 3 m / s and v q f = 1 . 1 m / s .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 17

phyexam3 - Version 061 Midterm 3 orin(58140 This print-out...

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

View Full Document
Ask a homework question - tutors are online