Question

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?

Answer 1

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

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

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

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

$\Rightarrow V_2=1.31 m/sec$

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

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

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