Question

In Fig., 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 μk is 0.500 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

Answer:

Explanation:

Given

initial height $h=2.5 m$

$m_2=2m_1$

coefficient of static friction $\mu =0.5$

When collision is elastic respective velocities after collision is

$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$

$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$

where $u_1$ and $u_2$=initial velocities of object

$v_1$ and $v_2$ final velocities of object

$u_1=\sqrt{2\times 9.8\times 2.5}$

$u_1=7 m/s$

$v_2=\frac{0+2m_1\times 7}{m_1+2m_1}$

$v_2=\frac{14}{3} m/s$

using $v^2-u^2=2 as$

$0-(4.67)^2=2\times (-0.5\times 9.8)\times s$

$s=2.22\ m$

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

$m_1u_1=(m_1+m_2)v$

$v=\frac{u_1}{3}=\frac{7}{3} m/s$

using $v^2-u^2=2 as$

$0-(2.33)^2=2\times (-0.5\times 9.8)\times s$

$s=0.55\ m$