Question

A block of mass m1 = 4 kg is moving at 4 m/s and collides with a block of mass m2 = 2 kg, which is moving at 5 m/s in the opposite direction on a frictionless floor. After the collision, the blocks stick together and move as a single unit. a) [5 points] Determine the total kinetic energy of the system before the collision.

Answer 1

Answer:

The total kinetic energy of the system before the collision is 57 Joules.                  

Explanation:

Given that,

Mass of first block, $m_1=4\ kg$

Speed of first block, $v_1=4\ m/s$

Mass of second block, $m_2=2\ kg$

Speed of first block, $v_2=-5\ m/s$

We need to find the total kinetic energy of the system before the collision. It is equal to the kinetic energies of both the blocks. It is given by :

$K_i=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2\\\\K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)\\\\K_i=\dfrac{1}{2}(4\times (4)^2+2\times (-5)^2)\\\\K_i=57\ J$

So, the total kinetic energy of the system before the collision is 57 Joules.