A prius with mass m1 (m1=1300 kg) traveling initially with a speed of v1 (=35 m/

Question

A prius with mass m1 (m1=1300 kg) traveling initially with a speed of v1 (=35 m/s) in an easterly direction crashes into the back of a U-haul truck (m2 = 10000kg) that was driving at a speed of v2 = 15 m/s

Find all:

1. calculate the speed of V

2. How much kinetic energy was last due to the crash?

3. Calculate the velocity of the Prius measured by the Truck driver after the collision (magnitude and direction)

4. Calculate the velocity of the Prius after the Collision (magnitude and Direction)

5 Calculate the velocity of the Truck after the collision (magnitude and Direction)

6 Using the speeds of Prius and the Truck, calculate the total kinetic energy of two after the collision.

Note that the error tolerance IS 2%. Show Work or explanat Hon to receive a Credit A Prius with mass ml (ml 300 kg) traveling initially with a speed of v1 (a 35.0 m/s) in an easterly direction e crashes into the back of a U-haul tnuck (m2 0,000 kg) that was driving at a speed of v2 5.0 m/s. (Q 1 and 2) Consider they got stuck after the crash and move together with a speed v I. Calculate the speed of V 2. How much kinetic energy was lost due to the crash? (Q 3-6) Consider the collision was elastic. Each is moving now with v1' and v2' 3. Calculate the velocity of the Prius measured by the Truck driver after the collision (magnitude and direction). 4. Calculate the velocity of the Prius after the collision (magnitude and direction). 5. Calculate the velocity of the Truck after the collision (magnitude and direction). 6. Using the specds of Prius and the Truck, calculate the total kinetic energy of two after the collision

Explanation / Answer

1.

m1 = 1300 kg

V1i = 35 m/s

m2 = 10000 kg

V2i = 15 m/s

V = velocity of combination after collision

using conservation of momentum

m1 V1i + m2 V2i = (m1 + m2) V

(1300) (35) + (10000) (15) = (1300 + 10000) V

V = 17.3 m/s

2)

Kinetic energy lost is given as

KElost = (0.5) [(m1 + m2) V2 - m1 V21i - m2 V22i ]

KElost = (0.5) [(1300 + 10000) (17.3)2 - (1300) (35)2 - (10000) (15)2]

KElost = - 2.3 x 105 J

3)

using conservation of momentum

m1 V1i + m2 V2i = m1 V1 + m2 V2

(1300) (35) + (10000) (15) = (1300 V1 + 10000 V2)

V1 = (195500 - 10000 V2)/1300 eq-1

using conservation of kinetic energy

m1 V21i + m2 V22i = m1 V21 + m2 V22

(1300) (35)2 + (10000) (15)2 = (1300 V21 + 10000 V22)

using eq-1

3.84 x 106 = 1300 ((195500 - 10000 V2)/1300)2 + 10000 V22

V2 = 19.6 m/s

in the east direction

using eq-1

V1 = (195500 - 10000 V2)/1300 = (195500 - 10000 (19.6))/1300

V1 = - 0.385 m/s

negative sign indicates west direction

Total Kinetic energy after collision is given as

KE = (0.5) (m1 V21 + m2 V22 )

KE = (0.5) (1300 (- 0.385)2 + 10000 (19.6)2)

KE = 1.92 x 106 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.