Question

In the figure, block 1 of mass m₁ slides from rest along a frictionless ramp from height h = 2.4 m and then collides with stationary block 2, which has mass m₂ = 2m₁. After the collision, block 2 slides into a region where the coefficient of kinetic friction μ_k is 0.2 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?

Frictionless 9

Answer 1

Height of ramp

Energy of block 1 at that height is (potential energy)

Energy of block 1 at the bottom of ramp is $E=\frac{1}{2}m_1u_1^2$

Conserving the energy of block 1 at two positions, $\frac{1}{2}m_1u_1^2=2.4m_1g\Rightarrow u_1=\sqrt{4.8g}=6.859\,m/s$

a) Elastic collision.

Velocity of block 1 before collision is   $u_1=6.859\,m/s$ and velocity of block 2 before collision is zero. $u_2=0\,m/s$

Conserving the momentum of the two blocks before and after collision,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+0=m_1v_1+(2m_1)v_2\,\,\Rightarrow \, v_1+2v_2=u_1$ ---(1)

For elastic collision $e=\frac{v_2-v_1}{u_1-u_2}=1$ $\Rightarrow v_2-v_1=u_1$ ----(2)

Solving (1) and (2) $v_2=\frac{2}{3}u_1=\frac{2}{3}*6.859=4.572\,m/s$ and $v_1=-2.287\,m/s$

With velocity $v_2=4.572\,m/s$ block $m_2$ travels a distance with friction coefficient $\mu_k=0.2$ until it stops.

Conserving the energy of mass $m_2$ ,

$-\mu_k m_2g d+\frac{1}{2}m_2v_2^2=0$

$d=\frac{v_2^2}{2\mu_k g}=\frac{(4.572)^2}{2*0.2*9.8}=5.33\,m$

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b) Inelastic collision

Velocity of block 1 before collision is   $u_1=6.859\,m/s$ and velocity of block 2 before collision is zero. $u_2=0\,m/s$

Conserving the momentum of the two blocks before and after collision,

$m_1u_1+m_2u_2=(m_1+m_2)v$

$m_1u_1+0=(m_1+2m_1)v\,\,\Rightarrow \, v=\frac{u_1}{3}$

Now both masses together travel a distance until they stop on the frictional surface.

Conserving the energy of mass $m_1+m_2$ ,

$-\mu_k (m_1+m_2)g d+\frac{1}{2}(m_1+m_2)v^2=0$

$d=\frac{v^2}{2\mu_k g}=\frac{u_1^2}{18\mu_k g}=\frac{(6.859)^2}{18*0.2*9.8}=1.33\,m$

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