Lesson_3.5_Homework_Solution

Lesson_3.5_Homework_Solution - Lesson 14 Homework Solutions...

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Lesson 14 Homework Solutions 1. When tossed upward and hit horizontally by a batter, a 0.32 kg ball receives an impulse of 5.0 Ns. With what speed does the ball move away from the bat? The initial horizontal velocity of the ball v i = 0 Initial horizontal momentum P i = 0 If v f is the horizontal velocity just after it was hit, the momentum after hitting P f = mv f = 0.32v f . The change in momentum = P f – P i = 0.32v f – 0 = 0.32v f The change in momentum = impulse 0.32v f = 5.0 Ns 1 5.0 0.32 15.6 f v ms == 2. A basketball with a mass of 0.6 kg is thrown horizontally against a wall with a velocity of 20 ms -1 . If the ball rebounds with a velocity of 15 ms -1 , what is the impulse imparted to the ball? The initial horizontal velocity of the ball v i = 20 ms -1 Initial horizontal momentum P i = 0.6 kg × 20 ms -1 = 12 kgms -1 The velocity of rebound v f = -15 ms -1 The final momentum P f = mv f = 0.6 kg (-15 ms -1 ) = -9.0 kg ms -1 The change in momentum = P f – P i = -9.0 – 12.0 = -19 kg ms -1 The change in momentum = impulse Impulse imparted to the ball = -19 Ns 3. A ball dropped from a height of 2.0 m rebounds to a height of 1.2 m after colliding with the floor. What is the coefficient of restitution between the floor and the ball? h 1 = 2.0 m and h 2 = 1.2 m Coefficient of restitution e is give by: 2 1 1.2 2 0.77 h e h =
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4. A freight car with a mass of 4.0 × 10 4 kg rolls down an inclined track a vertical distance of 4.2 m. At the bottom of the incline, on a level track, it collides and couples with an identical freight car that was at rest. What percentage of the initial kinetic energy is lost in the collision? The freight car at the top of the inclined track has a potential energy given by: U = mgh = 4.0 × 10 4 × 9.8 × 4.2 = 1.65 × 10 6 J All this energy become kinetic the bottom of the track. x If v is the velocity of this car at the bottom of the track, then: ½ mv 2 = mgh -1 2 2 9.8 4.2 9.1 ms vg h == × × = The momentum of this car just before colliding with the stationary car is: P 1i = mv = 4.0 × 10 4 kg × 9.1 ms -1 = 3.64 × 10 5 kg ms -1 Since the second car is at rest before collision, the total momentum before collision is: P i = 3.64 × 10 5 kg ms -1 If v f is the velocity of the two car after they collide and couple, then the total momentum of the system after collision is: P f = 2m v f = 8.0 × 10 4 v f Since P i = P f , 3.64 × 10 5 kg ms -1 = 8.0 × 10 4 v f v f = 4.55 ms -1 The total initial kinetic energy is: K i = ½ m 1 v 1i 2 + ½ m 2 v 2i 2 = ½ × 4.0 × 10 4 × 9.1 2 + 0 = 1.66 × 10 6 J The total kinetic energy after collision is: K f = ½ (m 1 + m 2 ) v f 2 = ½ × 8.0 × 10 4 × 4.55 2 = 8.3 × 10 5 J Kinetic energy lost = K i – K f = 8.3 × 10 5 J 5. A 1000-kg car traveling east at 72 kmh -1 and a 4000-kg SUV traveling south at 90 kmh -1 collide at a perpendicular intersection. Assuming a completely inelastic collision, what is the velocity of the vehicles immediately after collision?
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m 1 = 1000 kg, v 1i = 72 km h -1 = 20 ms -1 i , m 2 = 4000 kg, v 2i = 90 km h -1 = -25 ms -1 j Total momentum before collision is: P i = m 1 v 1i + m 2 v 2i = 1000 kg × 20 ms -1 i + 4000 kg × (-25 ms -1 j ) = 20,000 kg ms -1 i – 80,000 kg ms -1 j m 1 v 1i = 20,000 kg ms -1 i m 2 v 2i = – 80,000 kg ms -1 j 20,000 kg ms -1 i 80,000 kg ms -1 j Since P f = P i , we have P f = 20,000 kg ms -1 i – 80,000 kg ms -1 j But P f
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Lesson_3.5_Homework_Solution - Lesson 14 Homework Solutions...

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