m, A train car with mass m 682 kg is moving to the right with a speed of v1-7.2

Question

m, A train car with mass m 682 kg is moving to the right with a speed of v1-7.2 m/s and collides with a second train car. The two cars latch together during the collision and then move off to the right at - 4.4 m/s 1) What is the initial momentum of the first train car? 4910 kg-m/s Submit 2) What is the mass of the second train car? 434 kg Submit 3) What is the change in kinetic energy of the two train system during the collision? Submit Your submissions: 6874.56 X Computed value: 6874.56 Submitted: Monday, February 26 at 11:57 PM Feedback: Watch your sign the system loses energy during the collision. 4) m2 Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v1 7.2 m/s and collides with the second train car that is now moving to the left with a velocity v2- 5.3 m/s before the collision. The two cars latch together at impact. What is the final velocity of the two-car system? (A positive velocitý means the two train cars move to the right -a negative velocity means the two train cars move to the left.) 0.95 m/s Submit MacBook Pro

Explanation / Answer

1) p1i = m1 v1i

= 682 x 7.2

= 4910.4 kg m/s

2) Applying momentum conservation,

4910.4 = (m2 + 682)(4.4)

m2 = 434 kg ....Ans

3) Ki = (682)(7.2^2)/2 = 17677.44 J  

Kf = (682 + 434)(4.4^2)/2 = 10802.88 J

delta(K) = Kf - Ki = - 6875 J

(minus sign because energy is decreased.)

4) applying momentum conservation,

(682 x 7.2) + (434 x -5.3 ) = (682 + 434) v

v = 2.34 m/s

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