Answer to Question #107146 in Classical Mechanics for Angelo James

Question #107146
Show the complete solution. Use the correct number of significant figures

In the figure, block 1 of mass m1 slides from rest along a friction less ramp from height h = 3.2 m and then collides with stationary block 2, which has mass m2 = 5 m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction μk is 0.3 and comes to a stop in distance d within that region. What is the value of distance d if the collision is

(a) elastic and
(b) completely inelastic?
1
Expert's answer
2020-04-02T10:48:41-0400

As per the given question,

h=3.2m

let the mass of the first block="m_1"

and mass of the second block"(m_2)=5m_1"

kinetic friction "(\\mu_k)=0.3"

stopping distance(d)=?

Applying the conservation energy

"m_1gh=\\dfrac{m_1 v^2}{2}"

"\\Rightarrow v=\\sqrt{2gh}"

"\\Rightarrow v=\\sqrt{2\\times9.8\\times3.2}"

"\\Rightarrow v=7.91 m\/sec"

a) Now applying the conservation of momentum,

let the velocity after the collision is "v_2" and "V"

"m_1 v=m_2V+m_1 v_2"

"\\Rightarrow m_1\\times 7.91 =5m_1 V+m_1 v_2"

"\\Rightarrow 7.91=5V+v_2---------(i)"

"e=\\dfrac{V-v_2}{v_1}"

e=1

"\\Rightarrow v_1=V-v_2"

"7.91=V-v_2------------(ii)"

Now, adding the equation (i) and (ii)


"\\Rightarrow 6V=15.82m\/sec"

"\\Rightarrow V=\\dfrac{15.82}{6}=2.63m\/sec"

now, retardation ="\\mu g=0.3\\times 9.8=2.94 m\/sec^2"

"\\Rightarrow v^2=V^2-2as"

"\\Rightarrow 0=2.63^2-2\\times 2.94\\times s"

"s=\\dfrac{2.63\\times 2.63}{2\\times 2.94}=1.18 m"

ii) If collision is inelastic, let the final velocity is V2

"\\Rightarrow m_1v=(m_1+m_2)V_2"


"\\Rightarrow V_2=\\dfrac{m_1V}{m_1+m_2}"


"\\Rightarrow V_2=\\dfrac{m_1\\times7.91}{6m_1}=\\dfrac{7.91}{6}=1.31m\/sec"

"\\Rightarrow V_2=1.31 m\/sec"

"v^2=V^2-2(\\mu g)s"

"\\Rightarrow 0=1.31^2-2\\times 2.94 \\times s"

"\\Rightarrow s=\\dfrac{1.31\\times 1.31}{2\\times 2.94}=0.29m"


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