Question

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the person is 54.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 15.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard.

I got the answer: -0.6454 after using the conservation of momentum, but it's still wrong. What is the correct answer?

Answer 1

e Mass of person m, = 54.0 Kg Mass of skateboard = m₂ = 4.10 Kg Mass of brick = M3 = 2.50 Kg speed of brick relative to skateboard = v= 15 m/s let speed of brick relative to ground V3 let recoil speed of person and skatebo and = IN = =) speed of brick relative to skateboard (speed of brick relative to ground) -(speed of skateboard relative to ground) V = Vz-V2 = 0 = V + V2 . person, skateboard brick are initially at rest So, initial speed of system = V = 0 V3 (mitm2 + m3) V₂ Vz Å Vi = 0 mom (Initial) (Final) [Not on scale) Take left negative direction and right positive direction. Linear entum mass x speed =) Initial momentum of system = bi =) bi =(m, tmz+m3) Vo = bi =(m + M2 +M3) (0) =) bi = 0 momentum of brick after being throwing = P, = M3 V3

2d Momentum . of skateboard and person -(m,+ m2) V₂ P2 = So final momentum of the system = M₂ V3 -(mitm2 ) Vz Pf b, tez -) Pf td A = (21+1) Em = td (zut 'ul en Emt 1 Em en(7ut'm) - Em ) +1 Eu 21 ((am +'W) - en((amit > Pf = Pf - A Em -(-m3 + (mi an lumpuscusu bi = pf Now, by conservation of linear hare Initial momentum of system = Final mom entum of system =) o=m₃v - (-M3+(m + m2) / =) (-mz +(m, + m2)) V₂ m₂ v By using values in , we get (-2.5 kg +(54.0 kg +4.10 kg) v2 = 2.50 kg * 15 m/s =) (55.6 Kg ) V2 = 37.5 Kg mis 37.5 Kg mis = V₂ = 0.67446 m/s 55.6 Kg V ñ 0.674 m/s So, recoil speed of person and skateboard is 0.674 mis Ps Speed Ĥ always positive - V2 -