# Kinetic Energy Question?

## Car A with a mass of 1000kg is travelling east at 25m/s when it collides with car B, 2000kg, travelling north at 15m/s. After the collision, the two cars stick together and have the same velocity. Assume that they slide without friction after collision. (a) What is the final velocity of the combined mass of cars (magnitude and direction)? (b) How much kinetic energy was lost in this process and where did it go?

Nov 28, 2016

Hi, have a look and try to evaluate it...I cannot finish it now!

#### Explanation:

have a look and try evaluate ${v}_{x}$ and ${v}_{y}$....you can then use trigonometry and Pythagoras to characterize $v$::

Nov 28, 2016

See below.

#### Explanation:

In the absence of external forces

$\Delta \vec{p} = 0$ so

${m}_{{b}_{1}} {\vec{v}}_{{b}_{1}} + {m}_{{b}_{2}} {\vec{v}}_{{b}_{2}} = {m}_{{a}_{1}} {\vec{v}}_{{a}_{1}} + {m}_{{a}_{2}} {\vec{v}}_{{a}_{2}}$

Here we have

${m}_{{b}_{1}} = 1000 , {\vec{v}}_{{b}_{1}} = 25 \hat{i}$
${m}_{{b}_{2}} = 2000 , {\vec{v}}_{{b}_{2}} = 15 \hat{j}$

${m}_{{a}_{1}} = 1000 , {\vec{v}}_{{a}_{1}} = {\vec{v}}_{a}$
${m}_{{a}_{2}} = 2000 , {\vec{v}}_{{a}_{2}} = {\vec{v}}_{a}$

so

$1000 \cdot 25 \hat{i} + 2000 \cdot 15 \hat{j} = \left(1000 + 2000\right) {\vec{v}}_{a}$

so

${\vec{v}}_{a} = \left(\frac{25}{3} \hat{i} + 10 \hat{j}\right)$

with

$\left\mid {\vec{v}}_{a} \right\mid = \frac{5 \sqrt{61}}{3} = 13.02$

$\angle {\vec{v}}_{a} = \arctan \left(10 / \left(\frac{25}{3}\right)\right) = {50.2}^{\circ}$

Kinetic energy before ${K}_{b} = \frac{1}{2} {m}_{{b}_{1}} {\left\mid {\vec{v}}_{{b}_{1}} \right\mid}^{2} + \frac{1}{2} {m}_{{b}_{2}} {\left\mid {\vec{v}}_{{b}_{2}} \right\mid}^{2}$

Kinetic energy after

$\frac{1}{2} \left({m}_{{b}_{1}} + {m}_{{b}_{2}}\right) {\left\mid {\vec{v}}_{a} \right\mid}^{2}$

so

$\Delta K = {K}_{a} - {K}_{b} = - 283333.$[J]