A railway carriage of mass 10000kg moving with a speed 15/s hits a stationary carriage of same mass. after the collision the carriage get coupled and ove together. what is their common velocity after collision?

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A railway carriage of mass 10000kg moving with a speed 15/s hits a stationary carriage of same mass. after the collision the carriage get coupled and ove together. what is their common velocity after collision?

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Answer 1 · 2018-01-27T14:54:14

$7.5 m s^{- 1}$ $7.5ms^-1$

Explanation:

So the formula for linear momentum, $ρ$ $rho$ , is $ρ = m v$ $rho=mv$ .

So before the collision, the momentum of the system is $10000 \cdot 15 = 150000 k g m s^{- 1}$ $10000*15=150000kgms^-1$ .

After the collision, the momentum, according to the law of conservation of momentum, remains the same. That is, $ρ_{1} = ρ_{2}$ $rho_1=rho_2$ , or $m_{1} v_{1} = m_{2} v_{2}$ $m_1v_1=m_2v_2$

Here, $v_{2} = \frac{m_{1} v_{1}}{m_{2}}$ $v_2=(m_1v_1)/m_2$

$m_{1} v_{1} = 150000$ $m_1v_1=150000$ , we know this from our prior calculations. Now, since the second carriage gets stuck to the first one, the mass of the system, $m_{2}$ $m_2$ , is equal to $20000 k g$ $20000kg$

$v_{2} = \frac{150000}{20000}$ $v_2=150000/20000$

$v_{2} = 7.5 m s^{- 1}$ $v_2=7.5ms^-1$

Answer 2 · 2018-01-27T20:54:38

$= 7.5 m/s$ $=7.5"m/s"$

Explanation:

Applying the Law of Conservation of Momentum, identify the given data and set-up the needed formula to determine the common velocity of the carriages after the collision.

$Momentum before (P_{before}) = Momentum after (P_{a f t e r})$ $"Momentum before" (P_("before"))="Momentum after"(P_(after))$

$w h e r e :$ $where:$

$Momentum(P) = mass(m) \times velocity(v)$ $"Momentum(P)"="mass(m)"xx"velocity(v)"$
$P = m v$ $P=mv$

$(P_{before}) :$ $(P_("before")):$
The masses and the velocities of the carriages before the collision

$m_{1} = 10, 000 k g; v_{1} = 15m/s$ $m_1=10,000kg; v_1="15m/s"$
$m_{2} = 10, 000 k g; v_{2} = 0m/s (at rest)$ $m_2=10,000kg; v_2="0m/s " ( "at rest")$

$(P_{a f t e r}) :$ $(P_(after)):$
The masses and the velocities of the carriages after the collision

$m_1=10,000kg; v_1'=?$
$m_2=10,000kg; v_2'=?$

Take note that after the collision, both carriages moved together of same direction with the moving carriage. So that, a common velocity has been identified as follows:

$v_1'=v_2'=v_c="common velocity"$

Now, set-up the equation and plug in values in both conditions: before and after the collisions as described below.

$P_("before")=P_(after)$

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

but:

$v_1'=v_2'=v_c$

$m_1v_1+m_2v_2=m_1v_c'+m_2v_c'$

$(10,000kg)("15m/s")+cancel((10,000kg)("0m/s"))0=(10,000kg)(v_c)+(10,000kg)(v_c)$

$150,000"kg.m/s"+0=(20,000kg)v_c$ ; divide both sides by 20,000kg to isolate the variable $v_c$

$v_c=(150,000cancel(kg)."m/s")/(20,000cancel(kg))$

$v_c=7.5"m/s"$

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A railway carriage of mass 10000kg moving with a speed 15/s hits a stationary carriage of same mass. after the collision the carriage get coupled and ove together. what is their common velocity after collision?

2 Answers

Explanation:

Explanation:

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