ps04_solns - Physics 1311 Spring 2010 PS #4 Solutions Page...

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Physics 1311 Spring 2010 PS #4 Solutions Page 1 of 10 Physics 1311 Problem Set #4 Solutions Problem 1: A 6 . 20 g bullet is fred horizontally at two blocks resting on a smooth tabletop, as shown in the top fgure. The bullet passes through the frst block, with mass 1 . 20 kg, and embeds itselF in the second, with mass 1 . 80 kg. Speeds oF 0 . 430 m/s and 1 . 06 m/s, respectively, are thereby imparted to the blocks, as shown in the bottom fgure. (a) Neglecting the mass removed From the frst block by the bullet, fnd the speed oF the bullet immediately aFter it emerges From the frst block. (b) ±ind the bullet’s original speed. m 1 m 2 1 vv 2 v 0 Throughout this set of collisions, the total momentum of the system will be constant. Initially only the bullet is moving at some unknown speed, so we write the initial momentum as p i = m b v 0 . After the bullet passes through the Frst block, the block is moving at a known speed v 1 and the bullet is moving at some speed v b that is slower than v 0 . The total momentum now is p f 1 = m b v b + m 1 v 1 . ±inally, when the bullet lodges itself in the second block, the total momentum consists of these two objects moving together at v 2 , plus the Frst block still moving at v 1 : p f 2 = m 1 v 1 +( m b + m 2 ) v 2 . Note that we have all the quantities for p f 2 , so we can use this and the equation for p f 1 to solve for the unknown v b : p f 1 = p f 2 = m b v b + m 1 v 1 = m 1 v 1 m b + m 2 ) v 2 = m b v b =( m b + m 2 ) v 2 v b = m b + m 2 m b v 2 = ± 1+ m 2 m b ² v 2 . Numerically, this is v b = ± 1 . 80 0 . 00620 ² (1 . 06 m/s) = 308 . 8 m/s . = 309 m/s . Now, using the fact that momentum is conserved in the Frst collision, we can solve for the bullet’s initial speed: p i = p f 1 = m b v 0 = m b v b + m 1 v 1 = v 0 = v b + m 1 m b v 1 .
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Physics 1311 Spring 2010 PS #4 Solutions Page 2 of 10 Using our previous result and the given numbers, v 0 = (308 . 8 m/s) + 1 . 20 0 . 00620 (0 . 430 m/s) = 392 m/s . Problem 2: Two pucks collide on an air hockey table. Puck A has a mass of 18 g and is initially traveling in the + x direction at 7 . 1 m/s. Puck B has a mass of 54 g and is initially at rest. After the pucks collide, puck A moves away at an angle of 60 above the + x axis, while puck B travels at an angle of 25 below the + x axis. (a) Calculate puck A’s Fnal speed. (b) Calculate puck B’s Fnal speed. We begin by setting up the equations for this problem. In this 2-D collision, momentum is conserved in each coordinate direction . In the x direction, the conservation of momentum equation can be written as p ix = p fx = m A v i = m A v A cos θ A + m B v B cos θ B . In the y direction, there was no momentum to start with, so there’s no Fnal momentum either: p ix = p = 0= m A v A sin θ A - m B v B sin θ B . We write the second term with a negative sign because puck B heads o± below the x axis. θ A θ B v v v B A i A B A B (a) We have two momentum conservation equations, and two unknowns. To Fnd v A , we Frst solve the y equation for v B (actually for m B v B ): m B v B sin θ B = m A v A sin θ A = m B v B = m A v A sin θ A sin θ B .
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This note was uploaded on 02/21/2010 for the course PHYS 1311 taught by Professor Wiegert during the Spring '10 term at University of Georgia Athens.

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ps04_solns - Physics 1311 Spring 2010 PS #4 Solutions Page...

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