need help on a physics problem???

Question

In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.600 and comes to a stop in distance d within that region.

(a) What is the value of distance d if the collision is elastic?
(b) What is the value of distance d if the collision is completely inelastic?

http://i175.photobucket.com/albums/w138/kacortez/9-66.gif

Answer 1

Interesting problem.

Well, consider the kinetic energy and momentum of block 1 as it reaches the bottom of the slope.

KE = 1/2(m1)v^2 = (m1)gh
v = sqrt(2gh) = sqrt(2 * 9.8 * 2.5) = 7 m/s

Thus, the kinetic energy is 24.5 * m1 J and the momentum is 7 * m1 kg*m/s.

(a) In an elastic collision, momentum and KE are conserved. So you have 2 equations for 2 unknowns (since this is a 1D collision) - masses can cancel out since m2 = 2 * m1
24.5 * m1 = 1/2(m1)(v1)^2 + 1/2(m2)(v2)^2
24.5 = 1/2 * v2^2 + v2^2
49 = (v1)^2 + 2 * v2^2

and

7 * m1 = (v1)(m1) + (v2)(m2)
7 = v1 + 2 * v2

Substitute to solve for v1 and v2:
49 = (v1)^2 + 2 * v2^2
49 = (7 - 2 * v2)^2 + 2 * v2^2
49 = 49 - 28 * v2 + 6 * v2^2
0 = 6 * v2^2 - 28 * v2 = 3 * v2^2 - 14 * v2
0 = v2(3 * v2 - 14)

v2 = 0 OR v2 = 14/3

0 makes no sense, so we get 14/3 m/s. Now, it hits the region with friction - we need the acceleration due to the frictional force and solve for d using kinematics

F = 0.6 * (m2 * g)
a = F / m2 = 0.6 * g = 5.88

v = v0 + at
0 = 14/3 m/s - 5.88t
t = 0.79

x = x0 + v0t + (1/2)at^2
x = 5.55 m = d


(b) Almost the same thing, but KE is not conserved.

7 * m1 = v2 * (m1 + m2) = v2 * (3 * m1)
7 = 3 * v2
v2 = 7/3

Now, use the same formula for calculating the distance as before but just change the initial velocity. Answer you get is:

d = 3.7 m

Hope that helps

Answer 2

First, calculate the speed of block 1 just before collision using energy
m*g*h=.5*m*v^2
v=sqrt(2*g*h)
In this case
v=sqrt(2*9.82*2.50)
v=7.01 m/s

now let's look at conservation of momentum to compute the speed of block 2 after collision

a) for elastic, energy is conserved
m1*7.01=m1*v1+2*m1*v2
or
7.01=v1+2*v2
and
.5*m1*7.01^2=.5*m1*v1^2+.5*2*m1*v2^2
or
7.01^2=v1^2+2*v2^2
solve for v2

and then use energy to compute
.5*2*m1*v2^2=2*m1*g*0.6*d

b) for perfectly inelastic, the two stick together
m1*7.01=3*m1*v
or
v=7.01/3
then
.5*3*m1*v^2=3*m1*g*0.6*d

j

Answer 3

the galaxies rotate at speeds inconsistent with their obvious mass is via the fact we do see all of it. i'm bearing on it as being the theoretical dark count. There are very sturdy proofs that shows that dark count exist. One the is the inconsistent velocity of and obvious mass. dark count makes up approximately 75% to 80% of the situation interior the Universe...