Two small blocks A and B have masses 0.36kg and 0.32kg respectively. Block B is stationary on an ice rink 17m from a boundry wall. Block 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 immediately after the collision is 2.4ms-1.
a) Find the speed with which B starts to move?
b) 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?
Please show all your working out. Thank you
2007-02-17 22:14:54 · 3 answers · asked by Leonidus 2 in Science & Mathematics ➔ Physics
block b speed is 6.3 and final speed is 5.2
distance is 17 m
i dont recall the motion laws but u need to find the time and use energy or impulse to find the friction cooffeciant
2007-02-17 23:28:02 · answer #1 · answered by koki83 4 · 0⤊ 0⤋
a) 0.36(8)+0.32(0)= 0.36(2.4)+0.32v
2.016=0.32v
v= 6.3ms^-1
b) Use v^2= U^2 +2as
5.6^2=6.3^2+2a(17)
a=-0.245
The coefficient is -0.245. I hope this helps. ;)
2007-02-17 22:54:53 · answer #2 · answered by Juni Mccoy 3 · 0⤊ 0⤋
a) (8)(0.36)+(0.32)(0)=(0.36)(2.4)+(0.32)(v)
2.88=0.864+0.32v
v=(2.88-0.864)/0.32
=6.3ms-1
2007-02-17 22:53:12 · answer #3 · answered by apocalypse7489 2 · 0⤊ 0⤋