8. 0/5 points | Previous Answers MI4 10.6.021 My Notes A car of mass 2300 kg col

Question

8. 0/5 points | Previous Answers MI4 10.6.021 My Notes A car of mass 2300 kg collides with a truck of mass 4300 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was m/s, and the truck's velocity just before the collision was m/s (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. What system and principle should you use? Car plus truck Momentum principle Energy principle Car alone Truck alone (b) What is the velocity of the stuck-together car and truck just after the collision? v = | ?2121, 0, 20.5652) X m/s (c) In your analysis in part (b), why can you neglect the effect of the force of the road on the car and truck? The road doesn't exert forces on the car or truck and doesn't affect the vehicles. Short collision time, negligible impulse compared to large impulse acting between car and truck. (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)? AE (e) What kind of collision is this? inelastic elastic

Explanation / Answer

mc = 2300 kg

mt = 4300 kg

uc = ( 38 i + 0 j+0 k ) m/s

ut = ( -20 i + 0 j + 22 k ) m/s

a) car + truck system

and momentum principle

b) as the collission is inelastic so the after collission velocity will be same for the truck and car let it be

v = x i + y j z k

from momentume conservation

mc uc + mt ut = (mc+mt) v

2300  ( 38 i + 0 j+0 k ) + 4300 ( -20 i + 0 j + 22 k ) = (2300 + 4300) (x i + y j z k)

1400 i + 0 j + 94600 k = 6600 ( x i + y j z k)

v = x i + y j z k = 0.21 i + 0 j + 14.33 k

v = <0.2121 , 0 , 14.33>

c) Answer a

d) increased energy = kf - ki

= 1/2 x (2300+4300) x (14.2217)2 - [ 1/2 x 2300 x 382 + 1/2 x 4300 x 29.732 ]

= 677799.83 - [ 1660600 + 1900600 ]

= - 2.88 x106 J

answer

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