work power and forces

Question

The ravens safety Chuck Clark has a mass of 93kg. During a game, Clark tackles the 83 kg wide reciever for the steelers, Diontae Johnson. Chuck was running in a positive direction with a velocity of 8.12m/s and Johnson was running in the opposite direction with a velocity of 8m/s just before clark tackled him. Assume it was a perfectly inelastic collision.

a) what was chuck clarks momentum prior to the collision?

b) what will their velocity be after the collision?

c) what was chuck clarks momentum after the collision?

d) what impulse was exerted on chuck clark during the collision?

Answer 1

a) Momentum equals mass multiplied by velocity. So Chuck Clark has an initial momentum of (93)(8.12) = 755.16 kg•m/s, or, rounded to the appropriate number of sig figs, 760 kg•m/s.

b) A perfectly inelastic collision means they players stick together after colliding. But momentum is still conserved. Let P_chuck and P_diontae represent the initial momentums of Chuck and Diontae and let P_final represent the final momentum of the combined (stuck-together) players.

Since momentum is conserved:

P_chuck + P_diontae = P_final

m_chuck•v_chuck + m_diontae•v_diontae = (m_chuck + m_diontae)v_final

(93)•(8.12) + (83)(-8) = (93 + 83)v_final

755.16 - 664 = 176v_final

91.16 = 176v_final

v_final = 0.51796 or when rounded appropriately v_final = 0.52 m/s in the same direction that Clark was running

c) Chuck Clark's momentum after the collision was his mass multiplied by his (their) velocity = (93)(0.51796) = 48.1698 or, when rounded appropriately, P_chuck-final = 48 kg•m/s

d) Impulse equals change in momentum So the impulse was P_chuck-final - P_chuck = 48.1698 - 755.16 = -707 Ns (notice that although the units of kg•m/s could be used, that we typically use units of N•s or Ns for Impulse). The negative just means the force applied on him was in the opposite direction as his direction of travel.