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Unformatted text preview: garcia (jg46766) – Problem Set 8 – lacosse – (S2107) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Momentum Problem Set. 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 2 cm, at the maximum height the pendulum sub tends an angle of 29 . 9 ◦ , the mass of the bullet is 91 g, and the mass of the pendulum bob is 939 g. The acceleration of gravity is 9 . 8 m / s 2 . 939 g 91 g v i v f 2 9 . 9 ◦ 2 cm Determine the initial speed of the projec tile. Correct answer: 7 . 08662 m / s. Explanation: Let : θ = 29 . 9264 ◦ , m 1 = 91 g , m 2 = 939 g , and h = 2 cm = 0 . 02 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 (91 g) + (939 g) (91 g) bracketrightbigg × radicalBig 2 (9 . 8 m / s 2 ) (0 . 02 m) = 7 . 08662 m / s . 002 10.0 points Two particles of masses m and 6 m move to ward each other along the x − axis with the same initial speeds of 5 . 75 m / s. Mass m is traveling to the left and mass 6 m to the right. They undergo a headon elastic collision, and each rebounds. The motion through the en tire collision is entirely onedimensional. Find the final speed of the heavier particle. Correct answer: 2 . 46429 m / s. Explanation: Let : v = 5 . 75 m / s . Since the collision is elastic, we are allowed to conserve both momentum and kinetic en ergy. We take the first particle to have N times the mass of the second one (which has mass m ). First, we conserve momen tum. Then, we solve for the final velocity of the smaller mass, which we will then have to eliminate in the energy conservation equation. N mv − mv = N mv 1 f + mv 2 f ( N − 1) v = N v 1 f + v 2 f v 2 f = ( N − 1) v − Nv 1 f Now, we conserve the kinetic energy of the system, substituting the above expression in the last step, and then simplify. garcia (jg46766) – Problem Set 8 – lacosse – (S2107) 2 1 2 N mv 2 + 1 2 mv 2 = 1 2 N mv 2 1 f + 1 2 mv 2 2 f ( N + 1) v 2 = N v 2 1 f + v 2 2 f = N v 2 1 f + ( ( N − 1) v − Nv 1 f ) 2 = N v 2 1 f + ( N − 1) 2 v 2 + N 2 v 2 1 f − 2 N ( N − 1) v v 1 f 0 = N ( N + 1) v 2 1 f − 2 N ( N − 1) v v 1 f + v 2 bracketleftBig ( N − 1) 2 − ( N + 1) bracketrightBig = ( N + 1) v 2 1 f − 2 ( N − 1) v v 1 f + ( N − 3) v 2 This formula is quadratic in v 1 f , having the form 0 = a v 2 1 f + bv 1 f + c . We know that this....
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This note was uploaded on 03/28/2011 for the course PHYSICS 101 taught by Professor Carter during the Spring '11 term at University of Texas.
 Spring '11
 carter
 Physics, Momentum

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