Question

Pug the Rock-hopper penguin has a speed of
8.00 m/s at point A. His mass is 4.00 kg. The hill
between points A and C is frictionless
a. Pug slides down the hill from point A to
point B. Calculate Pug’s speed at point B.Pug arrives at point C and has an inelastic collision with Zazza the Adelie penguin, who is
initially at rest and is 2.00 kg. After the collision Zazza is moving rightwards at 18.0 m/s.
b. Calculate the velocity of Pug after the collision.
c. Calculate the magnitude of the impulse on Zazza due to the collision.
d. Find the average force that Pug exerts on Zazza during the 0.200 s long collision.
e. How much energy is lost in the collision?

Answer 1

From mechanical energy conservation between point A and B (taking ground as zero reference potential energy) KA+UA = KB +UB imu+mgha = amv, +mgh V3 = Vo+29(ha - hp) so velocity of the pug at point B is V V82 + (2) X (9.8) X (10-0) Vo = 16.125 x m/s (Zazza] » 23 Coral I omis Pug 4kg ea 16.125mls 2kg 4:0 4pf from momentum conservation (4) X (16.125) + (2) X(0) = (4) x (1)+(2) (18) V=7.125 x m/s Pug's velocity after collision V= 7.125 x m/s Impulse on zazza I = AP= m(V; -1) I = (2) X (18-0) = 36kg ms-! Favg = + = 32 = 180N Energy lost in the collision is ki - kg [x (4) X (16.125)2 +0] - [1 *(4) (7.125)2 + 1x (2)x(18)] - - 13767.688