Car B rests at the bottom of a frictionless inclined plane. In order to travel a height of 0.6m and maintain a speed of 2 m/s at the end of the track it needs to start with 4 m/s.

a) If car B perceives the speed through a fully elastic impact from car A, with what speed does car A have to hit car B if the mass of car A is twice the mass of car B?

b) What's the velocity of car A after the collision?

How I tried to solve it:

a)

$$m_A \cdot v_{A,End} = m_B \cdot v_{B, Beginning}$$

$$2m_B \cdot v_{A,End} = m_B \cdot 4 \frac{m}{s}$$

$$v_{A,End} = 2 \frac{m}{s}$$

Therefore $v_{A}$ has to be 2 $\frac{m}{s}$ when hitting car B.

b)

$$v_{A, End} = \frac{(m_A-m_B) \cdot v_{A, Beginning} + 2 m_B v_{B, Beginning}}{(m_A + m_B)}$$

$$v_{A, End} = \frac{(2m_B-m_B) \cdot 2 \frac{m}{s} + 2 \cdot m_B \cdot 4 \frac{m}{s}}{3 m_B}$$

$$v_{A, End} = \frac{m_B \cdot 2 \frac{m}{s} + 2 \cdot m_B \cdot 4 \frac{m}{s}}{3 m_B}$$

Which gives me 2 $\frac{m}{s}$ to the right.

According to the solutions, however, I should get $3 \frac{m}{s}$ for a) and $1 \frac{m}{s}$ for b).

• Why use pulleys when there are no ropes? – user80551 Jul 31 '13 at 8:38
• First remark: "if its mass is twice the mass of pulley A" means $m_B = 2m_A$. You've used $m_A = 2m_B$. Second remark: conservation of momentum should be applied by equating the sum of all the momenta before with that of all the momenta after the collision. – Wouter Jul 31 '13 at 8:41
• @user80551 my mistake, sorry... I thought pulleys are little cars as well. The question is not in English and I mistranslated it. + @ Wouter Typo. It should read "if mass A is twice the mass of B". I'll edit that. + apart from that; where exactly did I make a mistake? – libjup Jul 31 '13 at 8:48
• You'll also need the angle of inclination of the plane – user80551 Jul 31 '13 at 8:50
• @libjup Look at your equation for the conservation of momentum again. There are 4 momenta here: that of A before the collision, that of B before, that of A after and that of B after. You should put all the ones before on one side of the equation and the ones after on the other side, expressing that the total momentum is conserved by the collision. – Wouter Jul 31 '13 at 8:55

Let $u_a$ and $u_b$ be the final velocities of car a and b respectively, and let $v_a$ be the initially velocity of car a. The conservation of momentum that the final momentum is the same as the initial momentum (if there isn't any external forces): $$m_av_a=m_au_a+m_bu_b \tag1$$ Since this is an fully elastic impact, the energy is conserved: $$\dfrac{1}{2}m_av_a^2=\dfrac{1}{2}m_au_a^2+\dfrac{1}{2}m_bu_b^2 \tag2$$ With some algebra: $$m_av_a^2=m_au_a^2+m_bu_b^2\\ m_av_a^2-m_au_a^2=m_bu_b^2\\ m_a(v_a^2-u_a^2)=m_bu_b^2\\$$ Eq(2) becomes: $$m_a(v_a-u_a)(v_a+u_a)=m_bu_b^2 \tag3$$ Here one uses $a^2-b^2=(a-b)(a+b)$. Eq(1) can be rewritten as: $$m_a(v_a-u_a)=m_bu_b \tag4$$ Eq(4) can be inserted into Eq(3) to get: $$m_bu_b(v_a+u_a)=m_bu_b^2$$ or $$v_a+u_b=u_b \tag5$$ Plugging in that $m_a=2m_b$ and $u_b=4$ into Eq(1) and Eq(5) yields two equations: $$v_a+u_a=4 \tag6$$ $$2v_a=2u_a+4 \tag7$$ Eq(7) can be simplify to $$v_a=u_a+2 \tag8$$ Plugging in Eq(8) into Eq(6) gives: $$2u_a+2=4$$ which gives the result $u_a=1$ m/s. Plugging $u_a$ into Eq(8) gives $v_a=3$ m/s.