Question

HW 5.2.A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley whenit collides with and sticks to another 5.00-kg chunk of ice that is initially at rest. Since the valleyis icy, there is no friction. After the collision, how high above the valley floor will the combinedchunks go? (g= 9.8 m/s2)

Answer 1

The concept used to solve this problem is the conservation of momentum and the conservation of energy.

For conservation of the moment we have the definition:

$m_1v_1 = (m_1+m_2)v_f$

Where,

m = Mass

$v_1$ = Initial velocity for object 1

$v_f$ = Final velocity

Replacing the values we have to,

$m_1v_1 = (m_1+m_2)v_f$

$5*12=(5+5)v_f$

$v_f = 6m/s$

By conservation of energy we know that the potential energy is equal to the kinetic energy then

$mgh = \frac{1}{2} m(v_f^2-v_i^2)$

$gh = \frac{1}{2} v_f^2$

$h = \frac{1}{2} g*v_f^2$

$h = \frac{1}{2} (9.8)(6)^2$

$h = 1.837m$

Therefore after the collision the height when the combined chinks will go is 1.837m