Problem Set #8 Solutions

Problem 3 a this rst part is a collision problem in 1

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Unformatted text preview: ch means it doesn’t go as high as block 2. Problem 3: (a) This first part is a collision problem in 1-D, so we should use the principle of conservation of momentum. Since the blocks stick together, they have the same velocity vf immediately after the collision: pi = pf =⇒ m 1 v1 = ( m 1 + m 2 ) vf =⇒ vf = m1 v1 . m1 + m2 With the numbers, this is vf = 2. 5 2.6 m/s = 1.383 m/s ≃ 1.34 m/s. 2. 5 + 2. 2 (b) The second part of this problem can be solved using the idea of energy conservation. The kinetic energy of the blocks is converted entirely into spring potential energy, and This documentlost to friction. a JavaScript-enabled PDFthe collision, and no energy is requires So considering “initial” to be right after reader “final” to be when the spring is maximally compressed, such as Adobe Reader. Ef = Ei =⇒ Uspr,f = Ki . The kinetic energy needs to incorporate the motion of both of the blocks, using the value for vf that we calculated in part (a). Uspr,f = Ki 121 kx = (m1 + m2 )v 2 2 2 m1 + m2 2 x2 = v k m1 + m2 x= v k Now all that’s left is the numbers: x= 2. 5 + 2. 2 (1.383) = 0.2069 m ≃ 20.7 cm. 210 Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 3 of 9 Problem 4: Linear momentum is conserved in this collision, as in any collision where external forces can effectively be neglected. However, we don’t necessarily know that kinetic energy is conserved, so we won’t make that assumption. Since the second car is initially not moving, the initial momentum of our two-car system is simply pi = mv . After the collision, we can write the final momentum as pf = m(0.377v ) + (0.452m)vf where vf is the unknown final speed of the second car. Now we set pi = pf and solve for vf : pf = pi =⇒ m(0.377v ) + (0.452m)vf = mv vf = =⇒ 0.452mvf = mv − 0.377mv (1 − 0.377)mv = 1.38v. 0.452m Problem 5: We again want to use conservation of momentum in collisions as the central principle. We’re given enough information to find the final momentum of the two PDF reader This document requires a JavaScript-enabledstuck-together cars. . . but first we need to calculate their speed right after the collision. Sliding friction will exert a force on Adobe such as the cars of Reader. ff r = µk FN = µk Mtot g. This friction is the net force acting on the cars after the collision, so they experience a deceleration of magnitude Fnet = µk g. a= Mtot Knowing the distance they travel before stopping, constant-a...
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This note was uploaded on 08/15/2012 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at University of Georgia Athens.

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