Question

Rugby player A of mass of 120 kg running at 18 km/h jumps on another player B of mass 140 kg who is running in the same direction at 2.5 m/s. Find the common velocity considering rugby player B and player A, entangled after the collision.

Answer 1

$\bigstar\large\boxed{\mathtt{\green{Given:}}}$

• Mass of player A = 120 kg
• Initial velocity of player A = 18 km/hr.

In order to convert km /hr into m/s simply multiply by 5/18

Initial velocity of player A = 18 x5/18 = 5 m/s

• Mass of Player B = 140 kg
• Initial velocity of player B= 2.5 m/s

As both the players are moving in the same direction both the velocities will be positive and will move in the same direction with common velocity after collision

$\bigstar\large\boxed{\mathtt{\blue{To find:}}}$

• The common velocity of both the players after collision

$\bigstar\large\boxed{\mathtt{\orange{Solution:}}}$

As no external force is acting on the system the momentum of the system is conserved

Hence we can use law of conservation of momentum in order to solve this question ,

$\implies\mathtt{P_{initial}=P_{final}}$

$\implies\mathtt{m_av_a+m_bv_b=v(m_a+m_b)}$

$\implies\mathtt{120\times5+140\times2.5=v(120+140)}$

$\implies\mathtt{600+350=v(260)}$

$\implies\mathtt{950=v(260)}$

$\implies\mathtt{v=\dfrac{950}{260}}$

$\implies\mathtt{\red{v=3.7 \dfrac{m}{s}}}$

The common velocity of both the rugby players after collision is 3.7 m/s