Block A (mass mA = 1.15 kg) is sliding to the right on a horizontal, frictionles
ID: 1265410 • Letter: B
Question
Block A (mass mA = 1.15 kg) is sliding to the right on a horizontal, frictionless tabletop With a speed of va,0 = 2.4 m/s just before it collides with Block B. Block B (mass ma = 1.84 kg) Initially at rest. Block A collides with Block B. After the collision Block B moves to the right with a speed rat vb,f=1.75 m/s. (a) (5 Pts) How fast and in what direction is block A traveling after the collision? Answer: to the left to the right (circle one) (b) (4 pts) Next, Block B hits a portion of the table where there is friction. Block B comes to rest over a distance d = 1.40 m. What is the total work done by the frictional force acting on the block B? Answer: That is the magnitude of the total frictional force acting on the block B?Explanation / Answer
let, m1 = 1.15 kg, u1 = 2.4 m/s
m2 = 1.84 kg, u2 = 0
after the collsion, v2 = 1.75 m/s, v1 = ?
a) Apply momentum conservation,
m1*u1 + m2*u2 = m1*v1 + m2*v2
1.15*2.4 + 0 = 1.15*v1 + 1.84*1.75
==> v1 = (1.15*2.4 - 1.84*1.75)/1.15
= -0.4 m/s
b) Workdone by friction = chnage in kinetic enrgy
= 0.5*m2*(V3^2-v2^2)
= 0.5*1.84*(0 - 1.75^2)
= -2.82 J
c) W = F*d*cos(180)
-2.82 = -F*1.4
==> F = 2.82/1.4
= 2.01 N