Physics Solution Manual for 1100 and 2101

0 m2 v 2 cos 450 0 357 1 m1 v 1 cos 250

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Unformatted text preview: plies. The total linear momentum of the system, then, must remain the same before and after the rock is thrown. Let’s assume that m and v01 are, respectively the mass and initial velocity of the wagon, rider, and rock before the rock is thrown. In addition, let mrock be the mass of the rock, vf, rock be the final velocity of the rock after it is thrown, and vf be the final velocity of the wagon and rider (without the rock) after the rock is thrown. The conservation of linear momentum states that ( m − mrock 4f + mrock vf, rock = )v 14444 244444 3 mv 1201 44 3 Total momentum before rock is thrown Total momentum after rock is thrown Solving this equation for vf gives vf = m v01 − mrock vf, rock m − mrock SOLUTION In applying the momentum conservation principle, we assume that the forward direction is positive. Therefore the velocity of the wagon, rider, and rock before the rock is thrown is v01 = +0.500 m/s. Rock thrown forward: When the rock is thrown forward, its velocity is +16 m/s. The final velocity of the wagon is vf = m v01 − mrock vf, rock m − mrock = ( 95.0 kg ) ( +0.500 m/s ) − ( 0.300 kg ) ( +16.0 m/s ) = +0.451 m/s 95.0 kg − 0.300 kg Chapter 7 Problems 355 The final speed of the wagon (which is the magnitude of its velocity) is 0.451 m/s . Rock thrown backward: When the rock is thrown backward, its velocity is −16.0 m/s. In this case the final velocity of the wagon is vf = m v01 − mrock vf, rock m − mrock = ( 95.0 kg ) ( +0.500 m/s ) − ( 0.300 kg ) ( −16.0 m/s ) = +0.552 m/s 95.0 kg − 0.300 kg The final speed of the wagon is 0.552 m/s . 21. SSM REASONING The two-stage rocket constitutes the system. The forces that act to cause the separation during the explosion are, therefore, forces that are internal to the system. Since no external forces act on this system, it is isolated and the principle of conservation of linear momentum applies: m1 v f1 + m2 v f2 = ( m1 + m2 ) v 0 4 3 14 244 4 3 14 244 Total momentum after separation Total momentum before separation where the subscripts "1" and "2" refer to the lower and upper stages, respectively. This expression can be solved for vf1. SOLUTION v f1 = = Solving for vf1 gives ( m1 + m2 ) v 0 − m2 v f2 m1 2400 kg + 1200 kg ( 4900 m / s) − (1200 kg)(5700 m / s) 2400 kg = +4500 m / s Since vf1 is positive, its direction is the same as the rocket before the explosion . 22. REASONING During the breakup¸ the linear momentum of the system is conserved, since the force causing the breakup is an internal force. We will assume that the +x axis is along the original line of motion (before the breakup), and the +y axis is perpendicular to this line and points upward. We will apply the conservation of linear momentum twice, once for the momentum components along the x axis and again for the momentum components along the y axis. 356 IMPULSE AND MOMENTUM SOLUTION The mass of each piece of the rocket after breakup is m, and so the mass of the rocket before breakup is 2m. Applying the conservation of momentum theorem along the original line of motion (the x axis) gives mv1 cos 30.0° + mv2 cos 60.0° = 2mv0 14444 244444 { 4 3 or v1 cos 30.0° + v2 cos 60.0° = 2v0 (1) P0, x Pf ,x Applying the conservation of momentum along the y axis gives mv1 sin 30.0° − mv2 sin 60.0° = { 0 14444 244444 4 3 P0, y Pf , y or v2 = v1 sin 30.0° sin 60.0° (2) a. To find the speed v1 of the first piece, we substitute the value for v2 from Equation (2) into Equation (1). The result is v sin 30.0° v1 cos 30.0° + 1 cos 60.0° = 2v0 sin 60.0° Solving for v1 and setting v0 = 45.0 m/s yields v1 = 2v0 sin 30.0° cos 30.0° + cos 60.0° sin 60.0° = 2 ( 45.0 m/s ) sin 30.0° cos 30.0° + cos 60.0° sin 60.0° = 77.9 m/s b. The speed v2 of the second piece can be found by substituting v1 = 77.9 m/s into Equation (2): v sin 30.0° ( 77.9 m/s ) sin 30.0° v2 = 1 = = 45.0 m/s sin 60.0° sin 60.0° 23. SSM REASONING No net external force acts on the plate parallel to the floor; therefore, the component of the momentum of the plate that is parallel to the floor is conserved as the plate breaks and flies apart. Initially, the total momentum parallel to the floor is zero. After the collision with the floor, the component of the total momentum parallel to the floor must remain zero. The drawing in the text shows the pieces in the plane parallel to the floor just after the collision. Clearly, the linear momentum in the plane parallel to the floor has two components; therefore the linear momentum of the plate must be conserved in each of these two mutually perpendicular directions. Using the drawing in the text, with the positive directions taken to be up and to the right, we have Chapter 7 Problems x direction y direction − m1 v 1 (sin 25.0 ° ) + m2 v 2 (cos 45.0 ° ) = 0 357 (1) m1 v 1 (cos 25.0 ° ) + m2 v 2 (sin 45.0° ) – m3 v 3 = 0 (2) These equations can be solved simultaneously for the masses m1 and m2. SOLUTION Using the values given in the drawing for the velocities after th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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