Question

A 5-kg block(m1) is released (from rest) from the top of a 3.0-m high, frictionless incline. When the 5-kg block reaches the bottom of the incline it collides with a 3-kg block in a perfectly elastic collision. After the collision, the two blocks slide horizontally on a rough surface(μk=0.31) until they come to rest. How far does the 3-kg block slide before coming to rest?

Answer 1

To solve this problem we will apply the concepts related to the conservation of energy (Potential and kinetic), later we can consider the conservation of kinetic energy to finally apply the theory of kinematic equations of linear motion and find the velocity before reaching rest . From the conservation of energy we have to,

$\frac{1}{2} mv_1^2 = mgh$

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(3)}$

$v_1 = 7.66m/s$

From conservation of kinetic energy we have that

$KE_i = KE_f$

There is not Kinetic Energy at the beginning, then

$KE_f = 0$

$\frac{1}{2} m_2v_2^2 - \frac{1}{2} m_1v_1^2 = 0$

$\frac{1}{2} m_2v_2^2 = \frac{1}{2} m_1v_1^2$

Rearranging to find the velocity 2

$v_2 = \sqrt{\frac{m_1}{m_2}}v_1$

$v_2 = \sqrt{\frac{5}{3}}(7.66)$

$v_2 = 9.8m/s$

From kinetic equation we have that

$\Delta V = 2ad$

$9.899^2 = 2(0.31)(9.8)d$

$d = 16.12m$

Therefore the distance is 16.12m