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 B after the
collision?
(Unit = m/s)
Remember: right is +, left is -

Answer 1

Answer:

1.88 m/s

Explanation:

from elastic collision

KE before collision = KE after collision

Answer 2

The velocity of car B after the collision 1.5143 m/sec.

Given to us:

Mass of car A, $m_A= 281\ kg$

Mass of car B, $m_B= 209\ kg$

Initial velocity of car A, $u_A= +2.82\ m/sec$

Initial velocity of car B, $u_B= -1.72\ m/sec$

According to the question the collision is elastic, therefore the initial velocity and the final velocity can be given by,

$v_A=$ Final velocity of car A,

$v_B=$  Final velocity of car B,

$u_A+u_B=v_A+v_B\\v_A=v_B-(u_A+u_B)$

we also know that the momentum is always conserved, therefore

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B$

putting the vale of $v_A$,

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B\\\\m_Au_A+m_Bu_B=m_A[v_B-(u_A+u_B)]+m_Bv_B$

putting in the numerical values we have,

$m_Au_A+m_Bu_B=m_A[v_B-(u_A+u_B)]+m_Bv_B\\\\\ [ 281\times (+2.82) ]+ [ 209\times (-1.72) ]= 281[ v_B-(+2.82-1.72) ]+ [ 209\times v_B ]\\\\\ [792.42]+[-359.48]=281[v_B-1.1]+[v_B\times 209]\\\\432.94=281v_B-309.1+209v_B\\\\(432.94+309.1)=281v_B+209v_B\\\\v_B=\dfrac{(432.94+309.1)}{(281+209)}\\\\v_B=1.5143\ m/sec$

Hence, the velocity of car B after the collision 1.5143 m/sec.

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