PROBLEM Copyright McGraw-Hill Education. Permission required for reproduction or display. SOLUTION. v 1 = 4 km/hr = 1.
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1 PROLEM ,000 Mg ocea lier has a iitial velocity of 4 km/h. Neglectig the frictioal resistace of the water, determie the time required to brig the lier to rest by usig a sigle tugboat which exerts a costat force of 150 kn. m = 35,000 Mg = kg 3 F = N v 1 = 4 km/hr = m/s mv Ft = 0 (35 10 kg( m/s ( N t = 0 1 t = 59.6 s t = 4 mi19 s
2 PROLEM Two disks slidig o a frictioless horizotal plae with opposite velocities of the same magitude v 0 hit each other squarely. Disk is kow to have a weight of 6-lb ad is observed to have zero velocity after impact. Determie (a the weight of disk, kowig that the coefficiet of restitutio betwee the two disks is 0.5, (b the rage of possible values of the weight of disk if the coefficiet of restitutio betwee the two disks is ukow. Total mometum coserved: mv + mv = mv + mv ( m v + m( v = 0+ mv 0 0 Relative velocities: m v = 1 v m v v = e( v v 0 0 v = ev ( (1 Subtractig Eq. ( from Eq. (1 ad dividig by v 0, m m m 1 e= 0 = 1+ e m = m m 1 + e Sice weight is proportioal to mass, (a With W = 6 lb ad e = 0.5, (b With W = 6 lb ad e = 1, W W = (3 1 + e 6 W = W = 3.00 lb 1 + ((0.5 6 W = = lb 1 + ((1 With W = 6 lb ad e = 0, 6 W = = 6 lb 1 + ((0 Rage:.00 lb W 6.00 lb
3 PROLEM g ball is movig with a velocity of magitude 6 m/s whe it is hit as show by a 1-kg ball which has a velocity of magitude 4 m/s. Kowig that the coefficiet of restitutio is 0.8 ad assumig o frictio, determie the velocity of each ball after impact. t-directio: Total mometum coserved: all aloe: Mometum coserved: Replacig ( v ti ( i Eq. (1 efore fter v = 6 m/s ( v = (6(cos 40 = m/s ( v t= 6(si 40 = m/s v = ( v = 4 m/s ( v = 0 t m( v t + m( v t = m( v t + m( v t (0.6 kg( m/s + 0 = (0.6 kg( v t+ (1 kg( v t.314 m/s = 0.6 ( v + ( v (1 t t m( v t= m( v t = ( v t ( v = m/s ( t.314 = (0.6( ( v.314 = ( v t ( v = 0 t t
4 PROLEM (Cotiued -directio: Relative velocities: Total mometum coserved: [( v ( v ] e= ( v ( v [(4.596 ( 4](0.8 = ( v ( v = ( v ( v (3 m( v + m( v = m( v + m( v (0.6 kg(4.596 m/s + (1 kg( 4 m/s = (1 kg( v + (0.6 kg( v 1.44 = ( v + 0.6( v (4 Solvig Eqs. (4 ad (3 simultaeously, Velocity of : Velocity of : ( v = m/s ( v = 1.80 m/s ( v t ta β = ( v = β = 37. β + 40 = 77. v = ( (5.075 = 6.37 m/s v = 6.37 m/s v = 1.80 m/s
5 PROLEM girl throws a ball at a iclied wall from a height of 3 ft, hittig the wall at with a horizotal velocity v 0 of magitude 5 ft/s. Kowig that the coefficiet of restitutio betwee the ball ad the wall is 0.9 ad eglectig frictio, determie the distace d from the foot of the wall to the Poit where the ball will hit the groud after boucig off the wall. Mometum i t directio is coserved Coefficiet of restitutio i -directio mv si 30 = mv (5(si 30 = v t v t = ( vcos30 e = v t 1.5 ft/s (5(cos30 (0.9 = v v = ft/s Write v i terms of x ad y compoets Projectile motio ( v x 0 = v (cos30 vt (si 30 = 19.49(cos30 1.5(si 30 = ft/s ( v = v (si 30 + v (cos30 = 19.49(si (cos30 y 0 t = 0.57 ft/s 1 t y = y0 + ( v y 0t gt = 3 ft + (0.57 ft/s t (3. ft/s
6 PROLEM (Cotiued t, y = 0 = t 16.1 t ; t = s x = x + ( v t = (1.4098; x = ft 0 x 0 d = x 3cot 60 = ( ft (3 ft cot 60 = ft d = 13.5 ft
7 PROLEM kg block is movig with a velocity v 0 of magitude v 0 = m/s as it hits the 0.5-kg sphere, which is at rest ad hagig from a cord attached at O. Kowig that µ k = 0.6 betwee the block ad the horizotal surface ad e = 0.8 betwee the block ad the sphere, determie after impact (a the maximum height h reached by the sphere, (b the distace x traveled by the block. Velocities just after impact Total mometum i the horizotal directio is coserved: Relative velocities: Solvig Eqs. (1 ad ( simultaeously: (a Coservatio of eergy: mv + mv = mv + mv 0 + (1 kg( m/s = (0.5 kg( v + (1 kg( v 4= v + v (1 ( v v e= ( v v (0 (0.8 = v v 1.6 = v v ( v = 0.8 m/s v =.4 m/s 1 T1 = mv 1 V1 = 0 1 T1 = m(.4 m/s =.88m T V = 0 = m gh T1 + V1 = T + V.88 m + 0= 0+ m (9.81 h h = 0.94 m
8 PROLEM (Cotiued (b Work ad eergy: 1 1 T1 = mv 1 = m(0.8 m/s = 0.3m T = 0 U1 = Ff x = mknx = mxmgx = (0.6( m(9.81 x U1 = 5.886m x T + U = T : 0.3m 5.886m x= x = m x = 54.4 mm
9 PROLEM lock is released from rest ad slides dow the frictioless surface of util it hits a bumper o the right ed of. lock has a mass of 10 kg ad object has a mass of 30 kg ad ca roll freely o the groud. Determie the velocities of ad immediately after impact whe (a e = 0, (b e = 0.7. Let the x-directio be positive to the right ad the y-directio vertically upward. Let ( v x, ( v y, ( v x ad ( v y be velocity compoets just before impact ad ( v x,( v y,( v x, ad ( v those just after impact. y ispectio, y Coservatio of mometum for x-directio: ( v = ( v = ( v = ( v = 0 y y y y While block is slidig dow: 0+ 0 = m ( v + m ( v ( v = β ( v (1 x x x x Impact: 0+ 0 = m ( v + m ( v ( v = β ( v ( where β = m / m Coservatio of eergy durig frictioless slidig: Iitial potetial eergies: mgh for, 0 for. Potetial eergy just before impact: V 1 = 0 Iitial kietic eergy: T 0 = 0 (rest x x x x Kietic eergy just before impact: 1 1 T1 = mv + mv T + V = T + V mgh mv mv ( m m v 1 = m(1 + β v = + = + β gh gh v = ( v x = v = 1+ β 1+ β (3
10 PROLEM (Cotiued Velocities just before impact: v v gh = 1+ β gh = β 1+ β alysis of impact. Use Eq. ( together with coefficiet of restitutio. ( v ( v = e[( v ( v ] x x x x β( v ( v = e[( v + β( v ] x x x x ( v = ev ( (4 x x Data: m m = 10 kg = 30 kg h = 0. m g = 9.81 m/s 10kg β = = kg From Eq. (3, ((9.81(0. v = = m/s (a e = 0: ( v x = 0 ( v x = 0 v = 0 v = 0 (b e = 0.7: ( v x = (0.7( = m/s ( v x = ( ( = m/s v = 1.01 m/s v = m/s
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