Collision of Two Cars

[M. M. Fahad Joy]

Homework Statement

A car of 800 kg was going towards east with a velocity of 36km/h and an another car of 1600 kg was going towards west with a velocity of 54km/h. After collision, in which direction and with how much velocity will the cars go?

Homework Equations

$m_1u_1+m_2u_2 = (m_1+m_2)v$

The Attempt at a Solution

Here, we can take the first cars' mass as 1 and the second car as 2
Then,
$m_1=800kg$
$m_2=1600kg$
$u_1=36km/h$
$=\frac{36*10^3}{60*60}$
$=10m/s$
$u_2=54km/h$
$=\frac{54*10^3}{60*60}$
$=15m/s$

We know,
$m_1u_1+m_2u_2 = (m_1+m_2)v$
$Or, v = \frac{m_1u_1+m_2u_2}{m_1+m_2}$
$= \frac{800*10+1600*15}{800+1600}$
$= 13.33 m/s$ (ans.)

 

[M. M. Fahad Joy]

Homework Statement

A car of 800 kg was going towards east with a velocity of 36km/h and an another car of 1600 kg was going towards west with a velocity of 54km/h. After collision, in which direction and with how much velocity will the cars go?

Homework Equations

$m_1u_1+m_2u_2 = (m_1+m_2)v$

The Attempt at a Solution

Here, we can take the first cars' mass as 1 and the second car as 2
Then,
$m_1=800kg$
$m_2=1600kg$
$u_1=36km/h$
$=\frac{36*10^3}{60*60}$
$=10m/s$
$u_2=54km/h$
$=\frac{54*10^3}{60*60}$
$=15m/s$

We know,
$m_1u_1+m_2u_2 = (m_1+m_2)v$
$Or, v = \frac{m_1u_1+m_2u_2}{m_1+m_2}$
$= \frac{800*10+1600*15}{800+1600}$
$= 13.33 m/s$ (ans.)

[/B]

Can't we use LaTex here?

 

[haruspex]

Homework Statement

A car of 800 kg was going towards east with a velocity of 36km/h and an another car of 1600 kg was going towards west with a velocity of 54km/h. After collision, in which direction and with how much velocity will the cars go?

Homework Equations

$$m_1u_1+m_2u_2 = (m_1+m_2)v$$

The Attempt at a Solution

Here, we can take the first cars' mass as 1 and the second car as 2
Then,
$$m_1=800kg$$
$$m_2=1600kg$$
$$u_1=36km/h$$
$$=\frac{36*10^3}{60*60}$$
$$=10m/s$$
$$u_2=54km/h$$
$$=\frac{54*10^3}{60*60}$$
$$=15m/s$$

We know,
$m_1u_1+m_2u_2=(m_1+m_2)v$
##v = \frac{m_1u_1+m_2u_2}{m_1+m_2}##
$$= \frac{800*10+1600*15}{800+1600}$$ $$= 13.33 m/s$$ (ans.)[/B]

The LaTeX on this site needs double dollar signs front and back. Better, use double hash signs (#) so that it is not forced onto a new line. I've fixed it up in the copy above, but could not get one line to work.
As to your algebra, pay attention to signs. The cars were going in opposite directions.

 

[M. M. Fahad Joy]

The LaTeX on this site needs double dollar signs front and back. Better, use double hash signs (#) so that it is not forced onto a new line. I've fixed it up in the copy above, but could not get one line to work.
As to your algebra, pay attention to signs. The cars were going in opposite directions.

Thanks a lot for your help.
Could you tell me what the answer is?

 

[M. M. Fahad Joy]

Is it right?

 

[haruspex]

Is it right?
View attachment 226215 View attachment 226216

Yes, that is the right answer. You did not need to convert to m/s, though. Since the speeds are given kph you can answer in kph (24).

 

[M. M. Fahad Joy]

Yes, that is the right answer. You did not need to convert to m/s, though. Since the speeds are given kph you can answer in kph (24).

Thank you. I got it.

 

[neilparker62]

Much simpler to use the following formula (for perfectly inelastic collisions):

Δp = μΔv where μ is the reduced mass [m1 * m2 / (m1+m2) ] of the colliding objects and Δv is their relative velocity. Then p1(after) = p1(before) - Δp and you can divide by m1 to obtain the combined velocity.

p1 is the momentum of the heavier vehicle but you could similarly calculate p2(before) - Δp (this answer would be negative) and divide by m2. Same result.