Recent content by chrisbowe82
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C
Force exerted in inelastic collision
The way I worked out part a) was this (m1 u1) + (m2 u2) = (m1+m2)v (7000x8) + (5000x0) = (7000+5000) ie, the 7000kg truck moving at 8m/s and 5000kg van at 0m/s so 56,000 + 5000 = 12,000v v=61,000/12,000 v=5.08m/s after the collision so it gives me 5.08m/s as the velocity after the collision...- chrisbowe82
- Post #5
- Forum: Introductory Physics Homework Help
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C
Force exerted in inelastic collision
Well, the initial velocity of the van is 0m/s before the collision the 61,000 came from 7000kgx8m/s(the wagon) + 5000kgx0m/s(the truck) =61,000 So, I'm thinking now... F=mass of the wagon x final velocity (7000 x 5.08) - mass x initial velocity (7000x8) giving me f=35,560-56,000 /0.3 F=18760 N...- chrisbowe82
- Post #3
- Forum: Introductory Physics Homework Help
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C
Force exerted in inelastic collision
Hi, this is my first post, I've had a look at some examples to try and use them to help with my answer, but I've had no luck so far. I've just started doing physics and not spent much time on this board, so if my formatting/layout anything is a bit dodgy, then please tell me. I Would really...- chrisbowe82
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- Collision Force Inelastic Inelastic collision
- Replies: 5
- Forum: Introductory Physics Homework Help