Question

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s. What is the velocity of car A after the collision? Would greatly appreciate the actual answer!

Answer 1

Answer:

Hello There. ☆~\《--_^■^_--》\~☆ The final velocity of the car A is -1.053 m/s. For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Hope It Helps!~ ♡

ItsNobody~ ☆

Answer 2

The velocity of CAR after the collision is "1.881 m/s".

Given:

$M_A = 281 \ kg \\\\ v_A= 2.82 \frac{m}{s}\\\\M_B = 209 \ kg \\\\ V_B = 1.72 \frac{m}{s}$

To find:

velocity=?

Solution:

Using the formula for the velocity of CAR after the collision:

$\to V_A = \frac{V_A(M_A -M_B) +2 M_B V_B}{M_A+M_B}$

         $= \frac{2.82(281 - 209) +2 \times 209 \times 1. 72}{ 281+ 209}\\\\ = \frac{2.82(72) +718.96}{490}\\\\= \frac{203.04 +718.96}{490}\\\\= \frac{922.00}{490}\\\\= \frac{922}{490}\\\\=1.881 \ \frac{m}{s}$

Therefore, the answer is "$\bold{1.881\ \frac{m}{s}}$"

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