A 1580 kg car moving south at 13.3 m/s collides with a 2380 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.16 m/s to the north. Find the velocity of the 2380 kg car before the collision.
________m/s to the south
2007-11-10 14:47:58 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Concervation of momentum
m1V1-m2V2= (m1+m2)V
V2=((m1+m2)V+m1V1)/m2
V2= ((1580 +2380) 5.16 + 1580x 13.3)/2380
V2= 17.4 m/s
2007-11-10 15:49:10 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Total momentum before collision = Total momentum after collision.
Let the velocity towards north be taken as +ve
1580 x (- 13.3) + 2380 x ( v) = (1580 + 2380) 5.16
21014 + 2380 v = 20433.6
21014 - 20433.6 = - 2380 v
580.4 = -2380 v
v = 580.4 / -2380 = - 0.2439 m/s towards south
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2007-11-10 15:41:19 · answer #2 · answered by Joymash 6 · 0⤊ 0⤋
Conservation of momentum. Before the collision, you have (from p=mv) 1580*13.3 = 21 014 kgm/s going south and 2380*v going north, so the total momentum is:
P = -21014+2380v
After the collision, you have one mass, (1580+2380 = 3960 kg) going 5.16 north. Conservation of momentum tells us:
P(before) = P(after)
-21014+2380v = 3960*5.16
-21014+2380v = 20 433.6
And you can solve for v:
v = (20433.6 + 21014)/(2380)
v = 17.414958m/s
Of course its positive, we said south was negative, and this guys going north.
v= 17.41 m/s north or
v = -17.41 m/s south
2007-11-10 15:09:12 · answer #3 · answered by Anonymous · 1⤊ 0⤋
This is an in-elastic collision. So you can use the formula for inelastic colluision to solve the problem. The following site should help you solve the problem.
Tip: If the two masses stick to each other after collision, then it is inelastic collision.
If the two masses do not stick on to each other after collision or bounce in different directions then it is said to be elastic collsion.
http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html
Have a nice day!
Physics is fun!
2007-11-10 14:57:45 · answer #4 · answered by !şρÄЯķ! 2 · 0⤊ 0⤋
conservation of momentum
M1*(V1)ini + M2(V2)ini=M1*(V1)f+M2*(V2)f
(V1)f=(V2)f same because stuck
M1=1580,(V1)ini=13.3,M2=2380,(V2)ini=?
1580*13.3+2380*(V2)ini=2380*5.16+1580*5.16
2380*(V2)ini=[2380*5.16+1580*5.16]-1580*13.3
2380*(V2)ini=[12280.8+8152.8]-21014
(V2)ini=-580.4/2380
(V2)ini=-0.243
negative Meaning direction assumed wrong
and initial velocity of 0.243 m/s
2007-11-10 15:27:08 · answer #5 · answered by HeavyRain 4 · 0⤊ 0⤋
-19.583 m/s..-ve since it originally is moving towards north..
2007-11-10 15:01:53 · answer #6 · answered by Anonymous · 0⤊ 0⤋
im going to bed =]
2007-11-10 14:50:04 · answer #7 · answered by La UNiCA DeSii 3 · 1⤊ 1⤋