Question

Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (5.0 kg), initially at rest. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision?

Answer 1

Answer:

$\Delta K=-160.89\ J$

Explanation:

It is given that,

Mass of object A, $m_A = 2\ kg$

Initial speed of object A, $u_A=15\ m/s$

Mass of object B, $m_B = 5\ kg$

Initial speed of object A, $u_B=0$ (at rest)

Le V is the final speed when they lock and move with one common velocity. Using the conservation of momentum to find it.

$m_Au_A+m_Bu_B=(m_A+m_B)V$

$m_Au_A=(m_A+m_B)V$

$V=\dfrac{m_Au_A}{(m_A+m_B)}$

$V=\dfrac{2\times 15}{(2+5)}$

V = 4.28 m/s

Initial kinetic energy of the system is :

$K_i=\dfrac{1}{2}m_Au_A^2$

$K_i=\dfrac{1}{2}\times 2\times (15)^2$

$K_i=225\ J$

Final kinetic energy of the system is :

$K_f=\dfrac{1}{2}(m_A+m_B)V^2$

$K_f=\dfrac{1}{2}\times (2+5)\times (4.28)^2$

$K_f=225\ J$

$K_f=64.11\ J$

Let $\Delta K$ is the change in kinetic energy of the system after the collision. It is given by :

$\Delta K=K_f-K_i$

$\Delta K=64.11-225$

$\Delta K=-160.89\ J$

So, the change in kinetic energy of the system is 160.89 Joules.