Mitch and Paul are playing football. Mitch (mM= 90 kg) is running down the field

Question

Mitch and Paul are playing football. Mitch (mM= 90 kg) is running down the field at a speed of 6 m/s. Paul (mP= 80 kg) is chasing Mitch, running diagonally at a speed of 8 m/s with an angle = 25°. when they arrive at the same location, Paul tackles Mitch, and they slide along the ground together.

a.) What is the velocity (magnitude and direction, relative to Mitch’s initial direction of travel) of the two immediately after their collision? Friction can be considered negligible during the brief time of their impact.

b.) How far do they slide along the ground together before coming to rest, assuming a coefficient of friction of k= 0.6?

Explanation / Answer

here,

mM = 90 kg , mP = 80 kg

um = 6 m/s i

uP = 8 * ( cos(theta) i + sin(theta) j )

up = 7.25 i m/s + 3.38 j m/s

theta = 25 degree

a)

let their final speed be v

using conservation of momentum

mM* uM + mP * uP = (mM + mP) * v

90 * 6 m/s i + 80 * ( 7.25 i m/s + 3.38 j m/s ) = ( 90 + 80) * v

v = 6.59 i m/s + 1.59 j m/s

magnitude , |v| = sqrt(6.59^2 + 1.59^2)

|v| = 6.78 m/s

direction , theta = arctan(1.59/6.59) degree

theta = 13.5 degree

the final speed is 6.78 m/s and 13.5 degree to tthe initial direction of Mitch's

b)

uk = 0.6

accelration , a = - uk * g

a = - 5.88 m/s^2

let the distance travelled before stopping be r

|v|^2 - 0 = 2 * a * r

r = 3.91 m

the stopping distance is 3.91 m

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