Question

A railway wagon of mass 6 tonnes moving with a velocity of 18 km/h collides with a stationary wagon of mass 4 tonnes and both move together after collision. What is their combined velocity after collision?

answer fastt :)​

Answer 1

Provided that:

•️ Mass of railway wagon = 6 tonnes

•️ Initial velocity of railway wagon = 18 kmph

• Mass of stationary wagon = 4 tonnes

• Initial velocity of stationary wagon = 0

To calculate:

• Combined velocity of the system after the collision.

Solution:

• Combined velocity of the system after the collision = 10.8 kmph

Using concept:

• Conservation of law of momentum

• Formula to convert tonnes into kg

Using formula:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

${\small{\underline{\boxed{\sf{\rightarrow \: 1 \: tonne \: = 1000 \: kg}}}}}$

Required solution:

✡️ Firstly let us convert 6 tonnes into kilograms by using suitable formula!

→ 1 tonne = 1000 kg

→ 6 tonne = 6 × 1000 kg

→ 6 tonne = 6000 kg

• Henceforth, converted!

✡️ Now let us convert 4 tonnes into kilograms by using suitable formula!

→ 1 tonne = 1000 kg

→ 4 tonne = 4 × 1000 kg

→ 4 tonne = 4000 kg

• Henceforth, converted!

✡️ Now by using conservation of momentum formula let us solve this whole question!

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 6000(18) + 4000(0) = (6000+4000)v \\ \\ :\implies \sf 108000 + 0 = 10000v \\ \\ :\implies \sf 108000 = 10000v \\ \\ :\implies \sf \dfrac{108000}{10000} \: = v \\ \\ :\implies \sf \dfrac{108}{10} \: = v \\ \\ :\implies \sf v \: = 10.8 \: kmph$

• Henceforth, the combined velocity of the system after the collision is 10.8 kmph!