Physics Solution Manual for 1100 and 2101

2 ms 96 kg 090 ms 12 ms 20 reasoning friction is

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Unformatted text preview: kg ⋅ m / s I = G kg ⋅ m / s J 310 H K 16 ° 15. REASONING AND SOLUTION The excess weight of the truck is due to the force exerted on the truck by the sand. Newton's third law requires that this force be equal in magnitude to the force exerted on the sand by the truck. In time t, a mass m of sand falls into the truck bed and comes to rest. The impulse is F∆t = m( vf − v 0 ) so The sand gains a speed v0 in falling a height h so F= m( v f − v 0 ) ∆t 352 IMPULSE AND MOMENTUM ( ) v0 = 2 gh = 2 9.80 m/s 2 ( 2.00 m ) = 6.26 m/s The velocity of the sand just before it hits the truck is v0 = −6.26 m/s, where the downward direction is taken to be the negative direction. The final velocity of the sand is vf = 0 m/s. Thus, the average force exerted on the sand is m F = ( vf − v 0 ) = ( 55.0 kg/s ) [( 0 m/s ) − ( −6.26 m/s ) ] = +344 N ∆t 16. REASONING The total momentum of the two-cart system is the sum of their individual momenta. In part a, therefore, we will use Pf = m1 v f1 + m2 v f2 to calculate the system’s total momentum Pf when both carts are rolling. In order to find the initial velocity v01 of the first cart in part b, we need to know the system’s initial momentum P0 = m1 v 01 + m2 v 02 . According to the principle of conservation of linear momentum, this is equal to the system’s final momentum Pf if the system is isolated. Let’s see if it is. The attractive magnetic forces felt by the carts are internal forces and do not, therefore, affect the application of the conservation principle. Friction is negligible, and the external gravitational force and normal forces balance out on the level track. Therefore, the net external force on the system is zero, and the system is indeed isolated. We conclude that the system’s initial momentum P0 is identical to the final momentum Pf found in part a. SOLUTION a. We calculate the system’s final momentum directly from the relation Pf = m1 v f1 + m2 v f2 , taking care to use the algebraic signs indicating the directions of the velocity and momentum vectors: Pf = ( 2.3 kg ) ( +4.5 m/s ) + (1.5 kg ) ( −1.9 m/s ) = +7.5 kg ⋅ m/s b. The system’s initial momentum is the sum of the carts’ initial momenta: P0 = m1 v 01 + m2 v 02 . But the second cart is held initially at rest, so v02 = 0 m/s. Setting the initial momentum of the system equal to its final momentum, we have P0 = Pf = m1 v 01 . Solving for the first cart’s initial velocity, we obtain v 01 = Pf +7.5 kg ⋅ m/s = = +3.3 m/s m1 2.3 kg Chapter 7 Problems 353 17. REASONING AND SOLUTION The collision is an inelastic one, with the total linear momentum being conserved: m1v1 = (m1 + m2)V The mass m2 of the receiver is m2 = m1v1 V − m1 = (115 kg ) ( 4.5 m/s ) − 115 kg = 2.6 m/s 84 kg 18. REASONING Since friction between the disks and the air-hockey table is negligible, and the weight of each disk is balanced by an upward-acting normal force, the net external force acting on the disks and spring is zero. Therefore, the two-disk system (including the spring) is an isolated system, and the total linear momentum of the system remains constant. With v01 and vf1 denoting the initial and final velocities of disk 1 and v02 and vf2 denoting the initial and final velocities of disk 2, the conservation of linear momentum states that m1vf + m2vf = m1v01 + m2v02 14 1 2442 4 3 144 244 3 Total momentum after spring is released Total momentum before spring is released Solving this equation for vf2 yields the final velocity of disk 2. SOLUTION Using the fact that v01 = v02 = v0 (= +5.0 m/s), and remembering that disk 1 comes to a halt after the spring is released (vf1 = 0 m/s), we have vf 2 = = 19. m1v01 + m2v02 − m1vf 1 m1v0 + m2v0 − m1vf1 = m2 m2 (1.2 kg )( +5.0 m/s ) + ( 2.4 kg )( +5.0 m/s ) − (1.2 kg )( 0 m/s ) = 2.4 kg +7.5 m/s SSM REASONING Let m be Al’s mass, which means that Jo’s mass is 168 kg – m. Since friction is negligible and since the downward-acting weight of each person is balanced by the upward-acting normal force from the ice, the net external force acting on the twoperson system is zero. Therefore, the system is isolated, and the conservation of linear momentum applies. The initial total momentum must be equal to the final total momentum. SOLUTION Applying the principle of conservation of linear momentum and assuming that the direction in which Al moves is the positive direction, we find 354 IMPULSE AND MOMENTUM b gb b g gb gb b g g m 0 m / s + 168 kg − m 0 m / s = m 0.90 m / s + 168 kg − m −1.2 m / s 144444 2444444 4 3 14444444244444443 Initial total momentum Final total momentum Solving this equation for m, we find that 0 = m ( 0.90 m/s ) − (168 kg ) (1.2 m/s ) + m (1.2 m/s ) m= (168 kg ) (1.2 m/s ) = 96 kg 0.90 m/s + 1.2 m/s 20. REASONING Friction is negligible and the downward-pointing weight of the wagon and its contents is balanced by the upward-pointing normal force from the ground. Therefore, the net external force acting on the wagon and its contents is zero, and the principle of conservation of linear momentum ap...
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