Physics Solution Manual for 1100 and 2101

For either ball h0 0 m and when either ball has

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Unformatted text preview: electron and the stationary hydrogen atom experience an elastic head-on collision, we can employ Equation 7.8b to determine how vf,H is related to v0,e. SOLUTION According to Equation 7.8b, the final speed vf,H of the hydrogen atom after the collision is related to the initial speed v0,e of the electron by 2me vf,H = m + m H e v0,e Substituting this expression into the ratio of the kinetic energies gives KE hydrogen, after collision KE electron, before collison = 1 2 2 mH vf,H 1 2 2 me v0,e 2me mH m + m H e = me 2 The right hand side of this equation can be algebraically rearranged to give KE hydrogen, after collision KE electron, before collison m = H m e 2 m 1+ H me 2 2 2 −3 = (1837 ) 1 + 1837 = 2.175 × 10 Chapter 7 Problems +y 44. REASONING For use in our solution we define the following masses and initial speeds of the bullets: Mass Initial Speed −3 Bullet 1 m1 = 4.50 × 10 Bullet 2 m3 v03 m1 30.0° v01 30.0° m2 = 4.50 × 10−3 kg v02 = 324 m/s Bullet 3 kg v01 = 324 m/s m3 = ? 375 +x v02 v03 = 575 m/s m2 The drawing shows the bullets just after they are fired from the guns. They collide at the origin of the x, y axes and form a stationary lump. Assuming that the momentumconservation principle applies, we can conclude that the total momentum of the three-bullet system is the same after the bullets collide as before they collide. Applying this principle will allow us to determine the unknown mass. SOLUTION Applying the momentum-conservation principle separately in terms of the x and y components of the total momentum, we have x Component = m1v01 cos 30.0° − m3v03 cos 30.0° 1444442444443 0 14 3 24 x component of total momentum after collision y Component x component of total momentum before collision = −m1v01 sin 30.0° − m3v03 sin 30.0° + m2v2 14444444 24444444 3 0 14 3 24 y component of total momentum after collision y component of total momentum before collision Both the x and y components of the total momentum of the three-bullet system are zero after the collision, since the bullets form a stationary lump. The x component of the initial momentum of bullet 1 is positive and the y component is negative, because this bullet is fired to the right and downward in the drawing. The x and y components of the initial momentum of bullet 3 are negative, because this bullet is fired to the left and downward in the drawing. Either of the two equations presented above can be solved for the unknown mass m3. From the equation for the x component of the total momentum, we find that m3 = m1v01 v03 ( 4.50 ×10−3 kg ) (324 m/s ) = 2.54 ×10−3 kg = 575 m/s 45. SSM REASONING The two balls constitute the system. The tension in the wire is the only nonconservative force that acts on the ball. The tension does no work since it is perpendicular to the displacement of the ball. Since Wnc = 0 J, the principle of conservation of mechanical energy holds and can be used to find the speed of the 1.50-kg ball just before 376 IMPULSE AND MOMENTUM the collision. Momentum is conserved during the collision, so the principle of conservation of momentum can be used to find the velocities of both balls just after the collision. Once the collision has occurred, energy conservation can be used to determine how high each ball rises. SOLUTION a. Applying the principle of energy conservation to the 1.50-kg ball, we have 1 1 mv 2 + mgh = 2 mv 2 + mgh 2 14f 244f 140 2440 4 3 4 3 Ef E0 If we measure the heights from the lowest point in the swing, hf = 0 m, and the expression above simplifies to 2 1 1 mv f2 = 2 mv 0 + mgh0 2 Solving for vf , we have 2 v f = v 0 + 2 gh0 = ( 5.00 m / s) 2 + 2(9.80 m / s 2 )(0.300 m) = 5.56 m / s b. If we assume that the collision is elastic, then the velocities of both balls just after the collision can be obtained from Equations 7.8a and 7.8b: v f1 = F − m Iv m G +m J m HK 1 2 01 1 and 2 v f2 = F 2 m Iv G +m J m HK 1 01 1 2 Since v01 corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the quantity vf calculated in part (a). With the given values of m1 = 1.50 kg and m2 = 4 .60 kg , and the value of v 01 = 5.56 m / s obtained in part (a), Equations 7.8a and 7.8b yield the following values: v f1 = –2.83 m / s and v f2 = +2.73 m / s The minus sign in vf1 indicates that the first ball reverses its direction as a result of the collision. c. If we apply the conservation of mechanical energy to either ball after the collision we have 1 1 mv 2 + mgh = 2 mv 2 + mgh 2 14f 244f 140 2440 4 3 4 3 Ef E0 Chapter 7 Problems 377 where v0 is the speed of the ball just after the collision, and hf is the final height to which the ball rises. For either ball, h0 = 0 m, and when either ball has reached its maximum height, vf = 0 m/s. Therefore, the expression of energy conservation reduces to 12 gh f = 2 v 0 or hf = 2 v0 2g Thus, the heights to which each ball rises after the collision are 1.50 - kg ball hf = 4.60 - kg ball hf = 2 v0 2g 2 v0 2g = (2.83 m / s) 2 = 0.409 m 2 (9.80 m / s 2 ) = (2.73 m / s) 2 = 0.380 m 2 (9.80 m /...
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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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