Question #342670

Car B of mass 800kg is moving at a speed of and collides with car A which is stationary

After the collision both cars travel in the same direction as the initial direction of car B

After the collision car A moves at 1.3m/s

A) what is of car B after the collision

B) Calculate the impulse extended on car B by car A

C) State the impulse extended on car A by car B


1
Expert's answer
2022-05-19T05:43:15-0400

The law of conservation of momentum says

mBvB=mBvB+mAvAm_Bv_B=m_Bv_B'+m_Av_A'

a)

vB=vB(mA/mB)vAv_B'=v_B-(m_A/m_B)v_A'

b)

ΔpAB=mB(vBvB)=mAvA\Delta p_{AB}=m_B(v_B'-v_B)=-m_Av_A'

c)

ΔpBA=ΔpAB\Delta p_{BA}=-\Delta p_{AB}


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