Question

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.

Calculate the final velocity of the cars. (Note that since both cars have an initial velocity, you cannot use Equations 7.6a and b. You must look for other simplifying aspects.)
Magnitude (answer in m/s

Direction ° (counterclockwise from west is positive)

(b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)

Answer 1

???

Answer 2

force acting during collision is internal so momentum is conserve
so (initial momentum = final momentum) in both directions
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Let Vx is and Vy are final velocities of car in +x and +y direction respectively.
initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)
- 750*25 + 1150*0 = (750+1150)Vx
initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)
750*0 - 1150*5 = (750+1150)Vy
from here you can calculate Vx and Vy
so final velocity V is
\[V= \sqrt (V_{x}^{2} +V_{y}^{2})\]
and angle make from +ve x axis is
\[\theta = \tan ^{-1}(\frac{V _{y} }{ V _{x} })\]
kinetic energy loss in the collision = final KE - initial KE

Answer 3

Vx = -18750
Vy = 5750

Answer 4

i think both are negative.

Answer 5

Your velocities are awfully high.

In this problem, we have what is known as a perfectly inelastic collision. This is the case because both bodies stick together after the collision.

An inelastic collision means that MOMENTUM is conserved, while kinetic energy is lost.

We need to determine the total momentum before the collision and after the collision.

The best way to do this is to find the momentum of each car in the X and Y directions. (We'll use the coordinate plane outlined by gyanu, where +Y is North and +X is East.)

Fortunately, the cars are aligned nicely.

\[p_{1,y} = m_1 (-v_1)\]\[p_{1,x} = 0\]\[p_{2,x} = 0\]\[p_{2,y} = m_2 (-v_2)\]

Realize that car 1 has no momentum in the x component because it is traveling due south. Car 2 has no momentum in the y-component because it is traveling due west.

After collision, the two cars remain in contact, therefore momentum is conserved (as previously mentioned).

\[p_{f,x} = p_{1,x} + p_{2,x}\]\[p_{f,y} = p_{1,y} + p_{2,y}\]

We can find the final velocity as\[v_f = \sqrt{ \left ( p_{f,x} \over m_1 + m_2 \right)^2 + \left ( p_{f,y} \over m_1 + m_2 \right)^2 }\]
The angle of the two cars after the collision is determined by the tangent of their momentums in the x and y direction, as follows.

\[\theta = \tan^{-1} \left ( p_{f,y} \over p_{f,x} \right )\]

To find the energy lost during collision, we need to find the total energy of both cars before and after the collision. \[KE_1 + KE_2 \ne KE_f\]\[\Delta KE = KE_1 + KE_2 - KE_f\]

Answer 6

so did you solve it? :D

Answer 7

not yet please help

Answer 8

so you need to use momentum conservation right?
can u tell me what is the inital momentum of first car?

Answer 9

equation 1/2mv^2?
this equation right

Answer 10

no its this one
p=mv

Answer 11

car one with mass of 1150 is 5750
car two with mass of 750 is 18750

Answer 12

sorry man.. had to log out

ok.. momentum is vector right?

so to find out total momentum you cannot directly add numbers..
your fisrt momentum 5750 is due south and second one due West
so what is the total momentum? can u calculate?!