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Molodets [167]
3 years ago
13

A 1 170.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 800.0 kg t

ruck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east. (a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the car truck system in the collision? J (c) Account for this change in mechanical energy.
Physics
1 answer:
kykrilka [37]3 years ago
6 0

Explanation:

It is given that,

Mas of the car, m_1=1170\ kg

Initial speed of the car, u_1=25\ m/s

Mass of the truck, m_2=9800\ kg

Initial speed of the car, u_2=20\ m/s

Final speed of the car, v_1=18\ m/s

(a) It is a case of elastic collision. Let v_2 is the final velocity of the truck right after the collision. Using the conservation of linear momentum to find it :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{1170\times 25+9800\times 20-1170\times 18}{9800}

v_2=20.83\ m/s

(b) Initial kinetic energy is given by :

k_i=\dfrac{1}{2}(m_1u_1^2+m_2u_2^2)

k_i=\dfrac{1}{2}\times (1170\times (25)^2+9800\times (20)^2)

k_i=2325625\ J

Final kinetic energy is given by :

k_f=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

k_f=\dfrac{1}{2}\times (1170\times (18)^2+9800\times (20.83)^2)

k_f=2315595.61\ J

The change in mechanical energy of the car truck system in the collision:

\Delta K=k_f-k_i

\Delta K=2315595.61-2325625

\Delta k=-10029.39\ J

The loss in kinetic energy is 10029.39 Joules.

(c) The change in mechanical energy gets changed energy gets changed in the form of heat and light.

Hence, this is the required solution.

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P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

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