Physics Solution Manual for 1100 and 2101

530 kg 400 103 kg 0513 ms2 149 103 1 400 103

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: k = µ k FN , where µk is the coefficient of kinetic friction and FN is the magnitude of the normal force that acts on the two-car system. There are only two vertical forces that act on the system; they are the upward normal force FN and the weight (m1 + m2)g of the cars. Taking upward as the positive direction and applying Newton's second law in the vertical direction, we have FN – (m1 + m2 ) g = (m1 + m2 )a y = 0 , or FN = (m1 + m2 ) g . Therefore, f k = µk (m1 + m2 ) g , and we have x= (m1 + m2 )vf2 2 µ k (m1 + m2 ) g = vf2 2µk g = (8.9 m/s) 2 = 5.9 m 2(0.68)(9.80 m/s 2 ) 40. REASONING Considering the boat and the stone as a single system, there is no net external horizontal force acting on the stone-boat system, and thus the horizontal component of the system’s linear momentum is conserved: m1vf 1x + m2 vf2x = m1v01x + m2 v02x (Equation 7.9a). We have taken east as the positive direction. SOLUTION We will use the following symbols in solving the problem: m1 = mass of the stone = 0.072 kg vf1 = final speed of the stone = 11 m/s v01 = initial speed of the stone = 13 m/s m2 = mass of the boat vf2 = final speed of the boat = 2.1 m/s v02 = initial speed of the boat = 0 m/s Because the boat is initially at rest, v02x = 0 m/s, and Equation 7.9a reduces to m1vf 1x + m2 vf2x = m1v01x . Solving for the mass m2 of the boat, we obtain m2vf2x = m1v01x − m1vf1x or m2 = m1 ( v01x − vf1x ) vf2x 372 IMPULSE AND MOMENTUM As noted above, the boat’s final velocity is horizontal, so vf2x = vf2. The horizontal components of the stone’s initial and final velocities are, respectively, v01x = v01 cos 15° and vf1x = vf1 cos 12° (see the drawing). Thus, the mass of the boat must be m2 = m1 ( v01x − vf1x ) vf2 = vf1 v01x 12° vf1x 15° v01 Initial and final velocities of the stone ( 0.072 kg ) (13 m/s ) cos15o − (11 m/s ) cos12o 2.1 m/s = 0.062 kg 41. SSM REASONING The two skaters constitute the system. Since the net external force acting on the system is zero, the total linear momentum of the system is conserved. In the x direction (the east/west direction), conservation of linear momentum gives Pf x = P0 x , or ( m1 + m2 ) v f cosθ = m1 v 01 y Note that since the skaters hold onto each other, they move away with a common velocity vf. In the y direction, Pf y = P0 y , or p ( m1 + m2 ) v f sin θ = m2 v 02 These equations can be solved simultaneously to obtain both the angle θ and the velocity vf. p 1 SOLUTION a. Division of the equations above gives θ = tan −1 F v I = tan L kg)(7.00 m / s) O 73.0° m (70.0 = G v J M kg)(3.00 m / s) P m (50.0 HK N Q 2 −1 02 1 01 b. Solution of the first of the momentum equations gives vf = m1 v 01 ( m1 + m2 ) cosθ = (50.0 kg)(3.00 m / s) = 4.28 m / s (50.0 kg + 70.0 kg)(cos 73.0 ° ) 2 x θ P f Chapter 7 Problems 373 42. REASONING +355 m/s a. The conservation of linear Block 1 Block 2 momentum can be applied to this three-object system (the bullet and (a) Before collisions the two blocks), even though the second collision occurs later than the first one. Since there is no vblock 2 +0.550 m/s friction between the blocks and the horizontal surface, air resistance is negligible, and the weight of each mblock 2 = 1530 g mblock 1 = 1150 g block is balanced by a normal mbullet = 4.00 g force, the net external force acting on this system is zero, and the (b) After collisions conservation of linear momentum applies. This principle will allow us to determine the velocity of the second block after the bullet imbeds itself. b. The total kinetic energy of the three-body system is not conserved. Both collisions are inelastic, and the collision with block 2 is completely inelastic since the bullet comes to rest within the block. As with any inelastic collision, the total kinetic energy after the collisions is less than that before the collisions. SOLUTION a. The conservation of linear momentum states that the total momentum of the system after the collisions [see part (b) of the drawing] is equal to that before the collisions [part (a) of the drawing]: mblock 1vblock 1 + ( mblock 2 + mbullet ) vblock 2 = mbullet vbullet 14243 14444444 244444444 4 3 Total momentum Total momentum after collisions before collisions Solving for the velocity vblock 2 of block 2 after the collisions gives vblock 2 = mbullet vbullet − mblock 1vblock 1 mblock 2 + mbullet ( 4.00 × 10−3 kg ) ( +355 m/s) = − (1.150 kg )( +0.550 m/s ) 1.530 kg + 4.00 × 10−3 kg = +0.513 m/s b. The ratio of the total kinetic energy (KE) after the collision to that before the collision is 374 IMPULSE AND MOMENTUM KEafter = KE before = 1m v2 2 block 1 block 1 + 1 2 2 ( mblock 2 + mbullet ) vblock 2 1m v2 2 bullet bullet 1 2 (1.530 kg + 4.00 × 10−3 kg ) ( 0.513 m/s)2 = 1.49 × 10−3 1 4.00 × 10−3 kg 355 m/s 2 ) )( 2( (1.150 kg )( 0.550 m/s )2 + 1 2 43. REASONING The ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is KE hydrogen, after collision KE electron, before collison = 1 2 2 mH vf,H 1 2 2 me v0,e where vf,H is the final speed of the hydrogen atom, and v0,e is the initial speed of the electron. The ratio mH/me of the masses is known. Since the...
View Full Document

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.

Ask a homework question - tutors are online