Question

2. A 67 kg Samer is jogging down the street at a modest 8.2 m/s when he sees a 2 kg skateboard rolling in the same direction that he is running, just ahead of him. He could see that the skateboard was rolling with a constant velocity of 0.4m/s and decided to jump horizontally onto the top it. When he jumps onto the skateboard, Samer sticks without slipping, and the collision takes 0.15s. After the collision, the Samer-skateboard system continue in the same direction that they were originally moving.

(a) Determine the velocity of the Samer-skateboard system, after the collision and clearly state the type of collision that happened between Samer and the skateboard.

(b) Determine the average impulse on the skateboard, during the collision.

(c) the coefficient of static friction between Samer and the skateboard, during the collision. (Approximate the contact between Samer and skateboard to be a single point. You may ignore air resistance and inertial effects of the rotating wheels.) Hint: Draw an FBD that represents Samer or the skateboard during the collision.

(d) Determine the work done on Samer, due to the skateboard

Answer 1

Sol & m, = 2 kg M₂. = 67 kg, u = 0.4 nys, h = 8.2 nys a) If velocity after collision is then according to conservation of momentan (mitm) v = miu, + Mz Uz orv -mu, +4242 - 2x0.4 +67X8.2 - 7.97245 - mitru a b) Impuls; a change in momentun 2+67 hence i= my (2r-u) = 2x(7.97-0.4) = 15.15 N.S c) Here are force - frictional force of hon f- Hobits = 101N Now coefficient of static friction o olbar=-0.15 d) Here the work done on Samer by sketboard is zero because there is no relative motion between Sketboard and Samey,