# Tippens capitulo 9 ingles

Páginas: 6 (1443 palabras) Publicado: 26 de agosto de 2012
Chapter 9. Impulso y cantidad de movimiento

Choques elásticos e inelásticos
9-22.Un automóvil que circulaba a 8 m/s choca contra otro de la misma masa que estaba detenido frente a un semáforo. ¿Cuál es la velocidad de los autos chocados inmediata mente después de la colisión suponiendo que ambos se mantenga juntos?
( u1 = 8.00 m/s; u2 = 0, m1 = m2 = m)
mu1 + mu2 = (m + m)vc ; mu1 = 2mvc[pic] ; vc = 4.00 m/s
9-23. Un camion de 2000 kg que viaja a 10 m/s
m1u1 + m2u2 = (m1 + m2)vc ; (2000 kg)(10 m/s) + 0 = (2000 kg + 1200 kg)vc
20,000 m/s = 3200 vc ; vc = 6.25 m/s
½m1u12 + 0 =(m1 + m2)vc2; ½(2000 kg)(10 m/s)2 = ½(3200 kg)(6.25 m/s)2 + Loss
Loss = 100,000 J – 62,500 J; Loss = 37,500 J

9-24. A 30-kg child stands on a frictionless surface. Thefather throws a 0.8-kg football with a velocity of 15 m/s. What velocity will the child have after catching the football?
m1u1 + 0 = m1v1 + m2v2; (0.8 kg)(15 m/s) = (30 kg + 0.8 kg)vc
(30.8 kg)vc = 12 m/s; vc =0.390 m/s

*9-25. A 20-g object traveling to the left at 8 m/s collides head on with a 10-g object traveling to the right at 5 m/s. What is their combined velocity after impact?m1u1 + m2u2 = (m1 + m2)vc ; (20 g)(-8 m/s) + (10 g)(5 m/s) = (20 g + 10 g)vc
-110 m/s = 30 vc ; vc = -3.67 m/s, to left

*9-26. Find the percent loss of energy for the collision in Problem 9-25.
Conservation of Energy: ½m1u12 + ½m2u22 =½(m1 + m2)vc2 + Loss
½(20 g)(-8 m/s)2 + ½(10 g)(5 m/s)2 = ½(30 g)(-3.67 m/s)2 + Loss
765 J = 202 J + Loss; Loss = 563 J; [pic]%Loss =[pic]= 73.6%

*9-27. A 2-kg block of clay is tied to the end of a string as shown in
Fig. 9-9. A 500-g steel ball embeds itself into the clay causing both
to rise a height of 20 cm. Find the entrance velocity of the ball?
Before applying momentum conservation, we need to know the common velocity of the clay and ball after the collision. Energy is conserved : ½(m1 + m2) vc2 =(m1 + m2) gh
[pic]; vc = 1.98 m/s
m1u1 + 0 = (m1 + m2) vc ; (0.5 kg) u1 = (0.5 kg + 2 kg)(1.98 m/s)
(0.5 kg)u1 = 4.95 m/s; u1 = 9.90 m/s

*9-28. In Problem 9-27, suppose the 500-g ball passes entirely through the clay an emerges with a velocity of 10 m/s. what must be the new entrance velocity if the block is to raise to the same height of 20 cm?
We must find thevelocity v2 of the clay (m2) after collision:
½(m1 + m2) v22 = (m1 + m2) gh
[pic]; v2 = 1.98 m/s;
Momentum is conserved: m1u1 + 0 = m1v1 + m2v2;
(0.5 kg)u1 = (0.5 kg)(10 m/s) + (2 kg)(1.98 m/s); u1 = 17.9 m/s

*9-29. A 9-g bullet is embedded into a 2.0 kg ballistic pendulum (see Fig. 8-13). What was theinitial velocity of the bullet if the combined masses rise to a height of 9 cm?
½(m1 + m2) vc2 = (m1 + m2) gh
[pic]; vc = 1.33 m/s
m1u1 + 0 = (m1 + m2) vc ; (0.009 kg) u1 = (0.009 kg + 2 kg)(1.33 m/s)
(0.009 kg)u1 = 2.68 m/s; u1 = 297 m/s
*9-30. A billiard ball moving to the left at 30 cm/s collides head on with another ball movingto the right at 20 cm/s. The masses of the balls are identical. If the collision is perfectly elastic, what is the velocity of each ball after impact? (Consider right as +.)
Momentum: m1u1 + m2u2 = m1v1 + m2v2 Given: m1 = m2 = m, v1 = -30 cm/s, v2 = 0
m(-30 cm/s) + 0 = mv1 + mv2 ; v1 + v2 = (-30 cm/s) + (20 cm/s); v1 + v2 = -10 cm/s
Energy (e =1): v2 – v1 = u1 – u2 = (-30 cm/s) – (20 cm/s); v2 – v1 = - 50 cm/s
From second equation: v2 = v1 – 50 cm/s; Substituting this for v2, we obtain:
v1 + (v1 - 50 cm/s) = - 10 cm/s; and v1 = 20 cm/s, to right
And, v2 = v1 – 50 cm/s = (20 cm/s) – 50 cm/s; v2 = -30 cm/s, to left

9-31. The coefficient of restitution for steel is 0.90. If a...

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