Question

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1800 kg and was approaching at 5.00 m/s due south. The second car has a mass of 700 kg and was approaching at 21.0 m/s due west. (a) Calculate the final velocity of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)

Answer 1

Answer:

The final velocity of the cars is 6.894 m/s

The kinetic energy lost is 117,441 J

Explanation:

Using the conservation of the linear momentum :

$P_i = P_f$

Where $P_i$ is the inicial linear momentum and $P_f$ is the final linear momemtum.

The linear momentum is calculated by:

$P = MV$

where M is the mass and V is the velocity.

First we identify the directions of the velocity of both cars:

• the first car is moving in the axis y for the south direction, we will take this direction like positive.
• the second car is moving in the axis x to the west, we will take this direction like positive.

The conservation of the linear momentum is made on direction x, so:

$P_i = P_f\\M_{1}V_{1ix} +M_{2}V_{2ix} = V_{sx}(M_1+M_2)$

where $M_1$ is the mass of the car 1, $V_{1ix}$ is the car 1's  inicial velocity in the axis x,  $M_2$ is the mass of the car 2, $V_{2ix}$ is the car 2's  inicial velocity in the axis x and $V_{sx}$ is the velocity of the car 1 and car 2 after the collition.

The car 1 just move in the axis y so, it dont have horizontal velocity ($V_{1ix}$ = 0)

now the equation is:

$M_2V_{2ix} = V_{sx}(M_1 +M_2)$

Replacing the values, we get:

(700 kg)(21 m/s) = $V_{sx}$( 1800 kg + 700 kg)

solving for $V_{sx}$:

$V_{sx} = \frac{700(21)}{1800+700}$

$V_{sx}$ = 5.88 m/s

Now we do the conservation of the linear momentum on direction y:

$P_i = P_f\\M_{1}V_{1iy} +M_{2}V_{2iy} = V_{sy}(M_1+M_2)$

where $M_1$ is the mass of the car 1, $V_{1iy}$ is the car 1's  inicial velocity in the axis y,  $M_2$ is the mass of the car 2, $V_{2iy}$ is the car 2's  inicial velocity in the axis y and $V_{sy}$ is the velocity of the car 1 and car 2 together after the collition.

The car 2 just move in the axis x so, it don't have horizontal velocity ($V_{2iy}$ = 0)

now the equation is:

$M_1V_{1iy} = V_{sy}(M_1 +M_2)$

Replacing the values, we get:

$1800(5 m/s) = V_{sy}(1800 +700)$

solving for $V_{sx}$:

$V_{sy} = \frac{1800(5)}{1800+700}$

$V_{sy}$ = 3.6 m/s

Now, we have the two components of the velocity and using pythagorean theorem we find the answer as:

$V_f = \sqrt{5.88^2+3.6^2}$

$V_f =$ 6.894 m/s

Finally we have to find the kinetic energy lost, so the kinetic energy is calculated by:

K = $\frac{1}{2}MV^2$

so, ΔK = $K_f -K_i$

where $K_f$ is the final kinetic energy and $K_i$ is the inicial kinetic energy.

then:

$K_i = \frac{1}{2}(1800)(5)^2+\frac{1}{2}(700)(21)^2= 176,850 J$

$K_f = \frac{1}{2}(1800+700)(6.894)^2 = 59,409 J$

so:

ΔK =  59409 - 176850 = -117441 J