solution13

solution13 - fullwood(tf5349 – Homework Linear Momentum...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: fullwood (tf5349) – Homework Linear Momentum – Catala – (49033) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 5 cm, at the maximum height the pendulum sub- tends an angle of 48 . 2 ◦ , the mass of the bullet is 86 g, and the mass of the pendulum bob is 947 g. The acceleration of gravity is 9 . 8 m / s 2 . 947 g 86 g v i v f 4 8 . 2 ◦ 5 cm Determine the initial speed of the projec- tile. Correct answer: 11 . 8909 m / s. Explanation: Let : θ = 48 . 1897 ◦ , m 1 = 86 g , m 2 = 947 g , and h = 5 cm = 0 . 05 m . The final velocity is v f = radicalbig 2 g h. Using conservation of momentum m 1 v i = ( m 1 + m 2 ) v f , so v i = bracketleftbigg ( m 1 ) + ( m 2 ) m 1 bracketrightbigg v f = bracketleftbigg ( m 1 ) + ( m 2 ) m 1 bracketrightbigg radicalbig 2 g h = bracketleftbigg (86 g) + (947 g) (86 g) bracketrightbigg × radicalBig 2 (9 . 8 m / s 2 ) (0 . 05 m) = 11 . 8909 m / s . 002 (part 1 of 3) 10.0 points A 16.8 kg canoe moving to the left at 10 . 3 m / s makes an elastic head-on collision with a 4.5 kg raft moving to the right at 7.9 m/s. After the collision, the raft moves to the left at 22.9 m/s. Disregard any effects of the water. a) Find the velocity of the canoe after the collision. Correct answer:- 2 . 05 m / s. Explanation: Basic Concept: m c vectorv c,i + m r vectorv r,i = m c vectorv c,f + m r vectorv r,f Given: Let to the right be positive: m c = 16 . 8 kg v 1 ,i =- 10 . 3 m / s m r = 4 . 5 kg v 2 ,i = +7 . 9 m / s v 2 ,f =- 22 . 9 m / s Solution: v c,f = m c v c,i + m r v r,i- m r v r,f m c = v c,i + m r v r,i m c- m r v r,f m c =- 10 . 3 m / s + (4 . 5 kg)(7 . 9 m / s) 16 . 8 kg- (4 . 5 kg)(- 22 . 9 m / s) 16 . 8 kg =- 2 . 05 m / s fullwood (tf5349) – Homework Linear Momentum – Catala – (49033) 2 which is 2 . 05 m / s to the left. 003 (part 2 of 3) 10.0 points b) What is the total kinetic energy before the collision? Correct answer: 1031 . 58 J. Explanation: Basic Concept: KE = 1 2 mv 2 Solution: KE i = 1 2 m c v 2 c,i + 1 2 m r v 2 r,i = 1 2 (16 . 8 kg)(- 10 . 3 m / s) 2 + 1 2 (4 . 5 kg)(7 . 9 m / s) 2 = 1031 . 58 J 004 (part 3 of 3) 10.0 points c) What is the total kinetic energy after the collision?...
View Full Document

Page1 / 6

solution13 - fullwood(tf5349 – Homework Linear Momentum...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online