Physics Question (1 Viewer)

theKingPin · Jul 12, 2015

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 · Jul 12, 2015

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

Answer is 0.533.

InteGrand · Jul 12, 2015

theKingPin said:
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).$

Librah · Jul 12, 2015

InteGrand said:
$For elastic collisions, the following laws hold:$

$\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).

theKingPin · Jul 12, 2015

Thanks InteGrand. Great detailed answer.

Physics Question (1 Viewer)

theKingPin

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InteGrand

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Librah

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