# Physics Question (1 Viewer)

#### theKingPin

##### Member
Hi, I am stuck on a physics question. I can do the first part but my equations don't seem to work in part B. All help will be appreciated. BTW, use the hints they help a lot .
The url for the question: https://goo.gl/dCWPQ5

Thanks

#### Librah

I was going to give a solution, then i saw Integrand in the thread.

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#### InteGrand

##### Well-Known Member
Hi, I am stuck on a physics question. I can do the first part but my equations don't seem to work in part B. All help will be appreciated. BTW, use the hints they help a lot .
The url for the question: https://goo.gl/dCWPQ5

Thanks
$For elastic collisions, the following laws hold:$

$\underline{Conservation of kinetic energy (KE):} \text{total initial KE}=\text{total final KE}, i.e. \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 (since there are two bodies in each collision in this question). u represents an initial velocity, and v represents a final velocity.$

$\underline{Conservation of momentum:} m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$

$For each collision in the problem (i.e. A with B, then B with C), these two laws will provide us with two equations and we will have two unknowns each time (final velocities of each particle), which we will be able to solve for.$

$\underline{\textbf{First Collision}}$

$This is the collision between A and B. We have u_A = u>0, u_B=0, m_A = m, m_B = 2m, and we would like to find v_B>0.$

$Conservation of KE: \frac{1}{2}mu^2 + 0 = \frac{1}{2}mv_A^2 + \frac{1}{2}(2m)v_B^2 (since initial KE of particle B is 0)$

$\Rightarrow mu^2 = mv_A^2 + 2mv_B^2$

$\Rightarrow u^2 = v_A^2 + 2v_B^2 \ \ (1)$

$Conservation of momentum: mu+0=mv_A + (2m)v_B$

$\Rightarrow u=v_A + 2v_B \ \ (2)$

$Rearrange (2): v_A = u-2v_B.$

$Substitute in (1): u^2 = (u-2v_B)^2 + 2v_B^2$

$\Rightarrow u^2 = u^2 - 4uv_B + 4v_B^2 + 2v_B^2$

$\Rightarrow 6v_B^2 - 4uv_B=0 (simplifying and rearranging).$

$As v_B>0, this means 6v_B - 4u=0\Rightarrow v_B =\frac{2}{3}u.$

$\underline{\textbf{Second Collision}}$

$This is the collision between B and C. Using capital letters for velocities now (to avoid confusion with the previous collision), we have U_B =v_B= \frac{2}{3}u>0, U_C=0, m_B = 2m, m_C = 3m, and we would like to find V_C>0 (U's represent initial velocities for Collision 2, and V's represent final velocities).$

$Conservation of KE: \frac{1}{2}(2m)U_B^2 + 0 = \frac{1}{2}(2m)V_B^2 + \frac{1}{2}(3m)v_C^2 (since initial KE of particle C is 0)$

$\Rightarrow 2m\cdot \left(\frac{2}{3}u \right)^2 = 2mV_B^2 + 3mV_C^2$

$\Rightarrow \frac{8}{9} u^2 = 2V_B^2 + 3V_C^2 \ \ (3)$

$Conservation of momentum: (2m)U_B+0=(2m)V_B + (3m)V_C$

$\Rightarrow 2\times \frac{2}{3}u=2V_B + 3V_C$

$\Rightarrow \frac{4}{3}u=2V_B + 3V_C \ \ (4)$

$Rearrange (4): V_B=\frac{ \frac{4}{3}u - 3V_C}{2}.$

$Substitute in (3): \frac{8}{9} u^2 = 2\times \left(\frac{ \frac{4}{3}u - 3V_C}{2} \right)^2 + 3V_C^2$

$=\frac{1}{2}\left(\frac{16}{9}u^2-8uV_C+9V_C^2 \right)+3V_C^2$

$\Rightarrow \frac{8}{9} u^2 = \frac{8}{9} u^2 - 4uV_C + \frac{9}{2}V_C^2 + 3V_C^2$

$\Rightarrow \frac{9}{2}V_C^2 + 3V_C^2 - 4uV_C =0$

$\Rightarrow \frac{15}{2}V_C^2 - 4uV_C =0$

$\Rightarrow V_C \left( \frac{15}{2}V_C - 4u \right)=0$

$As V_C>0, this means \frac{15}{2}V_C - 4u=0\Rightarrow V_C = \frac{8}{15}u\approx 0.533 u. So the answer is \fbox{0.533} (no units).$

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#### Librah

$For elastic collisions, the following laws hold:$

$\underline{Conservation of kinetic energy (KE):} \text{total initial KE}=\text{total final KE}, i.e. \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 (since there are two bodies in each collision in this question). u represents an initial velocity, and v represents a final velocity.$

$\underline{Conservation of momentum:} m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$

$For each collision in the problem (i.e. A with B, then B with C), these two laws will provide us with two equations and we will have two unknowns each time (final velocities of each particle), which we will be able to solve for.$

$\underline{\textbf{First Collision}}$

$This is the collision between A and B. We have u_A = u>0, u_B=0, m_A = m, m_B = 2m, and we would like to find v_B>0.$

$Conservation of KE: \frac{1}{2}mu^2 + 0 = \frac{1}{2}mv_A^2 + \frac{1}{2}(2m)v_B^2 (since initial KE of particle B is 0)$

$\Rightarrow mu^2 = mv_A^2 + 2mv_B^2$

$\Rightarrow u^2 = v_A^2 + 2v_B^2 \ \ (1)$

$Conservation of momentum: mu+0=mv_A + (2m)v_B$

$\Rightarrow u=v_A + 2v_B \ \ (2)$

$Rearrange (2): v_A = u-2v_B.$

$Substitute in (1): u^2 = (u-2v_B)^2 + 2v_B^2$

$\Rightarrow u^2 = u^2 - 4uv_B + 4v_B^2 + 2v_B^2$

$\Rightarrow 6v_B^2 - 4uv_B=0 (simplifying and rearranging).$

$As v_B>0, this means 6v_B - 4u=0\Rightarrow v_B =\frac{2}{3}u.$

$\underline{\textbf{Second Collision}}$

$This is the collision between B and C. Using capital letters for velocities now (to avoid confusion with the previous collision), we have U_B =v_B= \frac{2}{3}u>0, U_C=0, m_B = 2m, m_C = 3m, and we would like to find V_C>0 (U's represent initial velocities for Collision 2, and V's represent final velocities).$

$Conservation of KE: \frac{1}{2}(2m)U_B^2 + 0 = \frac{1}{2}(2m)V_B^2 + \frac{1}{2}(3m)v_C^2 (since initial KE of particle C is 0)$

$\Rightarrow 2m\cdot \left(\frac{2}{3}u \right)^2 = 2mV_B^2 + 3mV_C^2$

$\Rightarrow \frac{8}{9} u^2 = 2V_B^2 + 3V_C^2 \ \ (3)$

$Conservation of momentum: (2m)U_B+0=(2m)V_B + (3m)V_C$

$\Rightarrow 2\times \frac{2}{3}u=2V_B + 3V_C$

$\Rightarrow \frac{4}{3}u=2V_B + 3V_C \ \ (4)$

$Rearrange (4): V_B=\frac{ \frac{4}{3}u - 3V_C}{2}.$

$Substitute in (3): \frac{8}{9} u^2 = 2\times \left(\frac{ \frac{4}{3}u - 3V_C}{2} \right)^2 + 3V_C^2$

$=\frac{1}{2}\left(\frac{16}{9}u^2-8uV_C+9V_C^2 \right)+3V_C^2$

$\Rightarrow \frac{8}{9} u^2 = \frac{8}{9} u^2 - 4uV_C + \frac{9}{2}V_C^2 + 3V_C^2$

$\Rightarrow \frac{9}{2}V_C^2 + 3V_C^2 - 4uV_C =0$

$\Rightarrow \frac{15}{2}V_C^2 - 4uV_C =0$

$\Rightarrow V_C \left( \frac{15}{2}V_C - 4u \right)=0$

$As V_B>0, this means \frac{15}{2}V_C - 4u=0\Rightarrow V_C = \frac{8}{15}u\approx 0.533 u. So the answer is \fbox{0.533} (no units).$

From
$\underline{Conservation of kinetic energy (KE):} \text{total initial KE}=\text{total final KE}, i.e. \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 (since there are two bodies in each collision in this question). u represents an initial velocity, and v represents a final velocity.$

$\underline{Conservation of momentum:} m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$

Might've been quicker to find a general solution for velocity. You get something like this after a bit of algebra, v_B=(2m)v_A/(m+2m) and similarly for v_C=2(2m)v_B/(2m+3m).