Question

A 110-kg rugby player is sliding on a muddy field at a speed of 2 m/s toward the right as a 82-kg player is slipping at a speed of 3 m/s toward the left. As the two players collide, they grab onto each other and slide away together as one.

What kind of collision was this?

Set up the conservation of linear momentum for this collision, and determine the shared velocity (magnitude & direction) of the players immediately after the collision.

Would the players' total kinetic energy before the collision be equal to their total kinetic energy after? yes/no

Answer 1

As the two players collide, they grab onto each other.

So, this is Completely In-elastic collision.

In Completely inelastic collision, after impact both players will travel together, So

Using momentum conservation:

Pi = Pf

m1*V1 + m2*V2 = (m1 + m2)*V

here,

m1 = mass of player 1 = 110 kg

V1 = initial velocity of player 1 = +2 m/s (taking right direction as +x axis)

m2 = mass of player 2 = 82 kg

V2 = initial velocity of player 2 = -3 m/s

V = final velocity of both players = shared velocity = ?

So,

V = (m1*V1 + m2*V2)/(m1 + m2)

V = (110*2 - 82*3)/(110 + 82)

V = -0.135 m/sec

V = 0.135 m/s towards left

Initial kinetic energy will be:

KEi = 0.5*m1*V1^2 + 0.5*m2*V2^2 = 0.5*110*2^2 + 0.5*82*3^2 = 589 J

Final kinetic energy will be:

KEf = 0.5*m1*V^2 + 0.5*m2*V^2 = 0.5*(110+82)*0.135^2 = 1.75 J

then, KEi KEf

Also, it is inelastic collision.

So, Player's total kinetic energy before the collision will not equal to their total kinetic energy after the collision.

Answer = No

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