1. Block A (mass= 1.2 kg) is sliding to the right on a horizontal, frictionless
ID: 1781608 • Letter: 1
Question
1. Block A (mass= 1.2 kg) is sliding to the right on a horizontal, frictionless tabletop with a speed of vA just before it collides with Block B. Block B (mass= 4.0 kg) is initially at rest. Block A collides with Block B and they stick together, traveling off at a speed of 2.0 m/s to the right just after the collision.
a) How fast is Block A traveling just before it hits Block B?
b) What is the total kinetic energy of the two blocks after the collision?
c) Next, the two blocks hit a portion of the table where there is friction. They come to rest over a distance d= 1.20 m. What is the total work done by the frictional force acting on the two blocks?
Explanation / Answer
mass of block A mA = 1.2 kg
mass of block B mB = 4 kg
before cllision
speed of block A = vA
speed of block B = vB = 0
after collision the two blocks stick together
speed of combined blocks = v = 2 m/s
from conservation of linear momentum
Lf = Li
mA*vA + mB*vB = (mA+mB)*v
(1.2*vA) + (4*0) = (1.2+4)*2
vA = 8.67 m/s <<<<----------ANSWER
===================
b)
total kinetic energy Ktot = (1/2)*(mA+mB)*v^2
Ktot = (1/2)*(1.2+4)*2^2 = 10.4 J
----------------
c)
work done = change in KE
Wf = Kf - Ki = 0 - (1/2)*(mA+mB)*v^2
W = -10.4 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.