Two small blocks A and B hav masses 0.36kg and 0.32kg respectively. Block B is stationary on an ice rink 17m from a boundary wall. Blaock A is moving at 8ms-1 at right angles to the boundry wall when it strikes B. Block A continues in the same direction, and its speed immedeiately after the collision is 2.4ms-1.
a) Given that B moves with constant retardation, and reaches the boundry wall with speed 5.6ms-1, find the coefficient of friction between B and the surface of the ice?
Block B rebounds from the wall in the reverse direction and comes to rest 4m from the wall. (you may asume that A and B do not collide again)
b) Find the change in B's Momentum as a result of its impact with the wall?
2007-02-19 04:13:45 · 2 answers · asked by Leonidus 2 in Science & Mathematics ➔ Physics
a) Conservation of momentum. v=B's speed immediately after the collision.
Before =after
0.36 x 8 = 0.36 x 2.4 + 0.32 x v
Calculate v
v=v1 and v2=5.6 in B's motion to the wall. Distance s=17m
(v2)^2 = (v1)^2 + 2as
Calculate a
b) Change in momentum = mom. before - mom. after
(mom. after is negative, being in the opposite direction.)
= 0.32 x 5.6 + 0.32 x 2.4
2007-02-19 04:37:43 · answer #1 · answered by DriverRob 4 · 2⤊ 0⤋
I wouldn't say hate so much as it just shows you that some people on this sight are just here to gain points and not actually getting any benefit from it. I'm here to learn about what people think about and to help if i can in any way,also ask the occasional question.
2016-05-24 09:20:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋