Question

A 195-kg rugby player running east with a speed of 4.00 m/s tackles a 95.0-kg opponent running north with a speed of 4.5 m/s. Assume the tackle is a perfectly inelastic collision. (Assume that the +x-axis points towards the east and the +y-axis points towards the north.)

(a) What is the velocity of the players immediately after the tackle?

magnitude	_____ m/s
direction	________ ° counterclockwise from the +x-axis

(b) What is the amount of mechanical energy lost during the collision?
______J

Answer 1

m₁ = mass of first player = 195 kg

m₂ = mass of second player = 95 kg

Along X-direction :

V_1ix = velocity of first player before collision = 4.4 m/s

V_2ix = velocity of second player before collision = 0 m/s

V_x = velocity of combination after collision

Along Y-direction :

V_1iy = velocity of first player before collision = 0 m/s

V_2iy = velocity of second player before collision = 4.5 m/s

V_y = velocity of combination after collision

Using conservation of momentum along X-direction ::

m₁ V_1ix + m₂ V_2ix = (m₁ + m₂) V_x

195 x 4.4 + 95 x 0 = (195 + 95) V_x

V_x = 2.96 m/s

Using conservation of momentum along Y-direction ::

m₁ V_1iy + m₂ V_2iy = (m₁ + m₂) V_y

195 x 0 + 95 x 4.5 = (195 + 95)V_y

V_y = 1.47 m/s

net velocity = V = sqrt (V_x² + V_y²) = sqrt (2.96² + 1.47²) = 3.305 m/s

= tan^-1(V_y /V_x ) = tan^-1(1.47/2.96) = 26.41

b)

initial total mechanical energy = (0.5) m₁ V_1x² + (0.5) m₂ V_2y² = (0.5) 195 (4)² + (0.5) 95 (4.5)² = 2521.875 J

Final total mechanical energy = (0.5) (m₁ + m₂) V² = (0.5) (195 + 95) (3.305)² = 1583.84 J

lost energy = 2521.875 - 1583.84 = 938.035 J