Can somebody please show me how to do this because I am stuck on this problem. Thank You!
A 1.5 kg steel ball and 10.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 3 kg block at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of friction between the block and shelf to be 0.09. How far does the block move before coming to rest? The acceleration due to gravity is 9.81 m/s^2.
Answer in units of m
A 1.5 kg steel ball and 10.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 3 kg block at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of friction between the block and shelf to be 0.09. How far does the block move before coming to rest? The acceleration due to gravity is 9.81 m/s^2.
Answer in units of m
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Step 1: Total Energy) Use conservation of total energy to find the final speed when the steel ball reaches its lower position. That speed should be sqrt(2gh) where h is 10.5m
Step 2: Conservation of Momentum and Kinetic Energy) In elastic collisions, you have conservation of momentum and conservation of kinetic energy.
(m1i)(v1i) + (m2i)(v2i) = (m1f)(v1f) + (m2f)(v2f)
(1/2)(m1i)(v1i)^2 + (1/2)(m2i)(v2i)^2 = (1/2)(m1f)(v1f)^2 + (1/2)(m2f)(v2f)^2
We have two equations with two unknowns, but if we tried to directly solve for v1f or v2f, it will be very complicated. With the right algebraic manipulations, though, v1f and v2f have the following equations:
v1f = [(m1 - m2)/(m1 + m2)](v1i) + [(2(m2))/(m1 + m2)](v2i)
v2f = [2(m1)/(m1+m2)](v1i) + [(m2 - m1)/(m1 + m2)](v2i)
Since you know that the block is at rest, its initial velocity is 0, so the terms on the right can cancel out.
v1f = [(m1 - m2)/(m1 + m2)](v1i)
v2f = [2(m1)/(m1+m2)](v1i)
But we're mainly interested in v2f
v2f = [2(m1)/(m1+m2)]sqrt(2gh)
Step 3: Friction and Newton's 2nd Law) Now that the block is moving with this velocity, you can find the net force on the block due to friction. That force is the coefficient multiplied by the normal force (or contact force)
F = -f = -un = -u(m2)g = (m2)a
a = -ug
Step 4: Kinematics) Now use the kinematics equations
x = (1/2)at^2 + (v0)t + x0
Using
vf = at + v0, we can eliminate t and we can solve for x to get
[(vf)^2 - (vi)^2]/(2a) + x0 = x
Setting x0 = 0, vi = [2(m1)/(m1+m2)]sqrt(2gh), a = -ug, and vf = 0, we can find the position x.
This is a good problem utilizing different concepts from different chapters of the book.
Step 2: Conservation of Momentum and Kinetic Energy) In elastic collisions, you have conservation of momentum and conservation of kinetic energy.
(m1i)(v1i) + (m2i)(v2i) = (m1f)(v1f) + (m2f)(v2f)
(1/2)(m1i)(v1i)^2 + (1/2)(m2i)(v2i)^2 = (1/2)(m1f)(v1f)^2 + (1/2)(m2f)(v2f)^2
We have two equations with two unknowns, but if we tried to directly solve for v1f or v2f, it will be very complicated. With the right algebraic manipulations, though, v1f and v2f have the following equations:
v1f = [(m1 - m2)/(m1 + m2)](v1i) + [(2(m2))/(m1 + m2)](v2i)
v2f = [2(m1)/(m1+m2)](v1i) + [(m2 - m1)/(m1 + m2)](v2i)
Since you know that the block is at rest, its initial velocity is 0, so the terms on the right can cancel out.
v1f = [(m1 - m2)/(m1 + m2)](v1i)
v2f = [2(m1)/(m1+m2)](v1i)
But we're mainly interested in v2f
v2f = [2(m1)/(m1+m2)]sqrt(2gh)
Step 3: Friction and Newton's 2nd Law) Now that the block is moving with this velocity, you can find the net force on the block due to friction. That force is the coefficient multiplied by the normal force (or contact force)
F = -f = -un = -u(m2)g = (m2)a
a = -ug
Step 4: Kinematics) Now use the kinematics equations
x = (1/2)at^2 + (v0)t + x0
Using
vf = at + v0, we can eliminate t and we can solve for x to get
[(vf)^2 - (vi)^2]/(2a) + x0 = x
Setting x0 = 0, vi = [2(m1)/(m1+m2)]sqrt(2gh), a = -ug, and vf = 0, we can find the position x.
This is a good problem utilizing different concepts from different chapters of the book.