Question

Lets say Iron Man is falling straight down at a rate of 52.7 m/s (about terminal velocity) and he has a mass of 279 kg. Hulk, who has a mass of 521 kg, jumps up at an angle and is traveling at 23.5 m/s at an angle of 50˚ right before catching Iron Man.

A) What is the final magnitude of the velocity of Hulk and Iron Man immediately after Hulk catches him?

(b) At what angle are they falling? Measure from the horizontal

Answer 1

Since there is no external force applied (as given that Speed of iron man is about terminal velocity), So using momentum conservation:

Pi = Pf

Since both Iron man and Hulk is locked onto each other after collision, So their final velocity will be same

m1*v1 + m2*v2 = (m1 + m2)*V

m1 = mass of Iron man = 279 kg and

m2 = mass of Hulk = 521 kg

v1 = initial velocity of Iron man = 52.7 m/sec towards South = (-52.7 j) m/sec

v2 = initial velocity of Hulk = 23.5 m/sec at 50 deg above the horizontal

Using components method:

v2 = v2x i + v2y j = 23.5*cos 50 deg i + 23.5*sin 50 deg j = (15.1055 i + 18.0020 j) m/s

Now Using given values:

279*(-52.7 j) + 521*(15.1055 i + 18.0020 j) = (279 + 521)*V

V = [-279*52.7 j + 521*15.1055 i + 521*18.0020 j]/(279 + 521)

V = (521*15.1055)/(279 + 521) i + (521*18.0020 - 279*52.7)/(279 + 521) j

V = 9.8375 i - 6.6553 j

Now final speed will be:

|V| = sqrt (9.8375^2 + (-6.6553)^2) = 11.877 m/sec

|V| = Magnitude of final velocity of Hulk + Iron Man = 11.9 m/s

Direction after collision will be:

Direction = arctan (Vy/Vx) = arctan (-6.6553/9.8375) = -34.079 deg

Direction = -34.1 deg = 34.1 deg below the horizontal (Use -ve sign since angle is below the horizontal)

Let me know if you've any query.