CJ57.P.003 REASONING AND SOLUTION According to the impulse-momentum theorem (see Equation 7.4), F t = mv

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1 Solution to HW#7 CJ57.CQ.003. RASONNG AND SOLUTON a. Yes. Momentum is a ector, and the two objects hae the same momentum. This means that the direction o each object s momentum is the same. Momentum is mass times elocity, and the direction o the momentum is the same as the direction o the elocity. Thus, the elocity directions must be the same. b. No. Momentum is mass times elocity. The act that the objects hae the same momentum means that the product o the mass and the magnitude o the elocity is the same or each. Thus, the magnitude o the elocity o one object can be smaller, or example, as long as the mass o that object is proportionally greater to keep the product o mass and elocity unchanged. CJ57.CQ.3. RASONNG AND SOLUTON a. No. The person oerhead jumps straight down and, thereore, applies only a ertical orce to the boat. Since riction and air resistance are negligible, no horizontal orce is applied to the boat. According to the impulse-momentum theorem, this means that the horizontal momentum o the boat cannot change. b. The speed o the boat decreases. Momentum is mass times elocity. The only way or the horizontal momentum o the boat to remain unchanged when the mass increases due to the presence o the jumper is or the magnitude o the boat s elocity (that is, the speed) to decrease. CJ57.P.003 RASONNG AND SOLUTON According to the impulse-momentum theorem (see quation 7.4), t m m 0, where is the aerage net orce. Taking the direction o motion (downward) as the negatie direction and soling or the aerage net orce, we obtain mc 0h b62.0 kgg 1.10 m / s ( 5.50 m / s) +165 N t 1.65 s where the plus sign indicates that the orce acts upward. CJ57.P.009. SSM WWW RASONNG The impulse applied to the gol ball by the loor can be ound rom quation 7.4, the impulse-momentum theorem: t m m 0. Only the ertical component o the ball's momentum changes during impact with the loor. n order to use quation 7.4 directly, we must irst ind the ertical components o the initial and inal elocities. We begin, then, by inding these elocity components. SOLUTON The igures below show the initial and inal elocities o the gol ball.

2 Beore impact Ater impact 0 cos sin 30.0 cos 30.0 sin we take up as the positie direction, then the ertical components o the initial and inal elocities are, respectiely, 0 cos 30.0 and cos Then, rom quation 7.4 the magnitude o the impulse is Since 45 m / s, 0 y 0 y t m( ) m ( cos 30.0 ) ( cos 30.0 ) y 0y 0 t 2m cos ( kg ) (45 m / s)(cos 30.0 ) 3.7 N s 0 CJ57.P.7. SSM RASONNG Batman and the boat with the criminal constitute the system. Graity acts on this system as an external orce; howeer, graity acts ertically, and we are concerned only with the horizontal motion o the system. we neglect air resistance and riction, there are no external orces that act horizontally; thereore, the total linear momentum in the horizontal direction is consered. When Batman collides with the boat, the horizontal component o his elocity is zero, so the statement o conseration o linear momentum in the horizontal direction can be written as (m 1 + m 2 ) Total horizontal momentum m Total horizontal momentum Here, m 1 is the mass o the boat, and m 2 is the mass o Batman. This expression can be soled or, the elocity o the boat ater Batman lands in it. SOLUTON Soling or gies m 1 (510 kg)(+11 m/s) +9.3 m/s m 1 + m kg + 91 kg The plus sign indicates that the boat continues to moe in its initial direction o motion; it does not recoil. CJ57.P.020. RASONNG AND SOLUTON irst, ind the initial speed o Adol by 2 using the conseration o mechanical energy, m g h m. 1 A A 2 A A

3 A 2gh 2(9.80 m / s 2 )( 065. m) 3. 6 m / s A Now use the conseration o momentum c m m to ind the elocity o A Ah d, so that ma m 120 kg ( 36. m/s) 55. m/s A 78 kg The conseration o mechanical energy allows us to ind the height reached by d, h b c m/s g m/s 2 2 g h 15. m CJ57.P.025. RASONNG AND SOLUTON The collision is an inelastic one, with the total linear momentum being consered. m 1 1 (m 1 + m 2 )V m 2 The mass m 2 o the receier is m 1 1 V - m 1 (115 kg)(4.5 m/s) 2.6 m/s kg 84 kg CJ57.P.029. RASONNG The elocity o the second ball just ater the collision can be ound rom quation 7.8b (see xample 7). n order to use quation 7.8b, howeer, we must know the elocity o the irst ball just beore it strikes the second ball. Since we know the impulse deliered to the irst ball by the pool stick, we can use the impulse-momentum theorem (quation 7.4) to ind the elocity o the irst ball just beore the collision. SOLUTON According to the impulse-momentum theorem, t m m 0, and setting 0 0 and soling or, we ind that the elocity o the irst ball ater it is struck by the pool stick and just beore it hits the second ball is t N s m kg m/s Substituting alues into quation 7.8b (with 9.09 m/s), we hae 2m 2 1 2m m 1 + m 2 m + m m/s CJ57.P.031. SSM WWW RASONNG The system consists o the two balls. The total linear momentum o the two-ball system is consered because the net external

4 orce acting on it is zero. The principle o conseration o linear momentum applies whether or not the collision is elastic. m + m m When the collision is elastic, the kinetic energy is also consered during the collision m + m m Total kinetic energy Total kinetic energy SOLUTON a. The inal elocities or an elastic collision are determined by simultaneously soling the aboe equations or the inal elocities. The procedure is discussed in xample 7 in the text, and leads to quations 7.8a and 7.8b. According to quation 7.8: 1 m m m + m and 2m 1 m m 2 + Let the initial direction o motion o the 5.00-kg ball deine the positie direction. Substituting the alues gien in the text, these equations gie kg ball kg ball kg 750. kg 5.00 kg kg 2(5.00 kg) 5.00 kg kg (2.00 m / s) m / s (2.00 m / s) m / s The signs indicate that, ater the collision, the 5.00-kg ball reerses its direction o motion, while the 7.50-kg ball moes in the direction in which the 5.00-kg ball was initially moing. b. When the collision is completely inelastic, the balls stick together, giing a composite body o mass m 1 + m 2 which moes with a elocity. The statement o conseration o linear momentum then becomes ( m + m ) m The inal elocity o the two balls ater the collision is, thereore,

5 m 1 (5.00 kg)(2.00 m / s) m + m 5.00 kg kg m / s CJ57.P.032. RASONNG The net external orce acting on the two-puck system is zero (the weight o each ball is balanced by an upward normal orce, and we are ignoring riction due to the layer o air on the hockey table). Thereore, the two pucks constitute an isolated system, and the principle o conseration o linear momentum applies. SOLUTON Conseration o linear momentum requires that the total momentum is the same beore and ater the collision. Since linear momentum is a ector, the x and y components must be consered separately. Using the drawing in the text, momentum conseration in the x direction yields m A 0A m A A (cos 65 )+m B B (cos 37 ) (1) while momentum conseration in the y direction yields Soling equation (2) or B, we ind that 0 m A A (sin 65 ) m B B (sin 37 ) (2) B m A A (sin 65 ) m B (sin 37 ) (3) Substituting equation (3) into quation (1) leads to a. Soling or gies A m A 0A m A A (cos 65 )+ m A A (sin 65 ) sin 37 (cos 37 ) A 0A sin 65 cos 65 + tan m/s sin 65 cos 65 + tan m/s b. rom equation (3), we ind that B (0.025 kg) (3.4 m/s) (sin 65 ) (0.050 kg) (sin 37 ) 2.6 m/s