Apply Momentum Conservation to Collisions - Physics
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What is the momentum of a particle of mass $m$ moving at velocity $v$ in 1D?
What is the momentum of a particle of mass $m$ moving at velocity $v$ in 1D?
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$p = mv$. Momentum equals mass times velocity in the chosen direction.
$p = mv$. Momentum equals mass times velocity in the chosen direction.
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What is the momentum conservation equation for a 1D collision in an isolated system?
What is the momentum conservation equation for a 1D collision in an isolated system?
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$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. Total momentum before equals total momentum after in isolated systems.
$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. Total momentum before equals total momentum after in isolated systems.
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What condition must be true for total momentum to be conserved during a collision?
What condition must be true for total momentum to be conserved during a collision?
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$F_{\text{ext,net}} = 0$ (impulse from external forces is zero). No external forces means the system's total momentum stays constant.
$F_{\text{ext,net}} = 0$ (impulse from external forces is zero). No external forces means the system's total momentum stays constant.
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What is the impulse-momentum theorem for an object in 1D?
What is the impulse-momentum theorem for an object in 1D?
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$J = \Delta p = m(v-u)$. Impulse equals change in momentum, which is mass times velocity change.
$J = \Delta p = m(v-u)$. Impulse equals change in momentum, which is mass times velocity change.
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What value of $e$ corresponds to a perfectly elastic collision?
What value of $e$ corresponds to a perfectly elastic collision?
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$e = 1$. Objects separate at same relative speed they approached.
$e = 1$. Objects separate at same relative speed they approached.
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What value of $e$ corresponds to a perfectly inelastic collision (objects stick)?
What value of $e$ corresponds to a perfectly inelastic collision (objects stick)?
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$e = 0$. Objects stick together with no relative motion after collision.
$e = 0$. Objects stick together with no relative motion after collision.
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What is the post-collision common velocity $v$ if two masses stick together in 1D?
What is the post-collision common velocity $v$ if two masses stick together in 1D?
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$v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$. Total momentum divided by total mass gives common velocity.
$v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$. Total momentum divided by total mass gives common velocity.
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What additional conservation law applies to an elastic collision besides momentum?
What additional conservation law applies to an elastic collision besides momentum?
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Kinetic energy is conserved: $K_i = K_f$. Elastic collisions conserve both momentum and kinetic energy.
Kinetic energy is conserved: $K_i = K_f$. Elastic collisions conserve both momentum and kinetic energy.
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What is the kinetic energy of a mass $m$ moving at speed $v$?
What is the kinetic energy of a mass $m$ moving at speed $v$?
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$K = \frac{1}{2}mv^2$. Energy of motion equals half mass times velocity squared.
$K = \frac{1}{2}mv^2$. Energy of motion equals half mass times velocity squared.
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Identify the sign convention rule for 1D momentum problems along a chosen axis.
Identify the sign convention rule for 1D momentum problems along a chosen axis.
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Choose + direction; velocities opposite it are negative. Consistent signs ensure momentum conservation equations work correctly.
Choose + direction; velocities opposite it are negative. Consistent signs ensure momentum conservation equations work correctly.
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Find the final velocity $v$ if $m_1 = 2,\text{kg}$ at $u_1 = 3,\text{m/s}$ hits $m_2 = 1,\text{kg}$ at rest and they stick.
Find the final velocity $v$ if $m_1 = 2,\text{kg}$ at $u_1 = 3,\text{m/s}$ hits $m_2 = 1,\text{kg}$ at rest and they stick.
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$v = 2,\text{m/s}$. $(2)(3) + (1)(0) = (2+1)v$ gives $v = 2$ m/s.
$v = 2,\text{m/s}$. $(2)(3) + (1)(0) = (2+1)v$ gives $v = 2$ m/s.
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Find the final velocity $v$ if $m_1 = 1,\text{kg}$ at $u_1 = 4,\text{m/s}$ hits $m_2 = 3,\text{kg}$ at $u_2 = 0$ and they stick.
Find the final velocity $v$ if $m_1 = 1,\text{kg}$ at $u_1 = 4,\text{m/s}$ hits $m_2 = 3,\text{kg}$ at $u_2 = 0$ and they stick.
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$v = 1,\frac{\text{m}}{\text{s}}$. $(1)(4) + (3)(0) = (1+3)v$ gives $v = 1$ m/s.
$v = 1,\frac{\text{m}}{\text{s}}$. $(1)(4) + (3)(0) = (1+3)v$ gives $v = 1$ m/s.
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Find $v_2$ if $m_1 = m_2$, $u_2 = 0$, and after collision $v_1 = 0$ (use momentum conservation).
Find $v_2$ if $m_1 = m_2$, $u_2 = 0$, and after collision $v_1 = 0$ (use momentum conservation).
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$v_2 = u_1$. Equal masses exchange velocities in this special case.
$v_2 = u_1$. Equal masses exchange velocities in this special case.
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Find the system total momentum $p$ for $m_1 = 2,\text{kg}$ at $+3,\text{m/s}$ and $m_2 = 1,\text{kg}$ at $-2,\text{m/s}$.
Find the system total momentum $p$ for $m_1 = 2,\text{kg}$ at $+3,\text{m/s}$ and $m_2 = 1,\text{kg}$ at $-2,\text{m/s}$.
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$p = 4,\text{kg}\cdot\text{m/s}$. $(2)(3) + (1)(-2) = 6 - 2 = 4$ kg·m/s.
$p = 4,\text{kg}\cdot\text{m/s}$. $(2)(3) + (1)(-2) = 6 - 2 = 4$ kg·m/s.
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Find the final speed $v$ if a $0.20,\text{kg}$ cart at $5,\text{m/s}$ sticks to a $0.30,\text{kg}$ cart at rest.
Find the final speed $v$ if a $0.20,\text{kg}$ cart at $5,\text{m/s}$ sticks to a $0.30,\text{kg}$ cart at rest.
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$v = 2,\text{m/s}$. $(0.20)(5) + (0.30)(0) = (0.50)v$ gives $v = 2$ m/s.
$v = 2,\text{m/s}$. $(0.20)(5) + (0.30)(0) = (0.50)v$ gives $v = 2$ m/s.
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Find the recoil speed $v_r$ of a $2,\text{kg}$ gun if it fires a $0.010,\text{kg}$ bullet at $400,\text{m/s}$ from rest.
Find the recoil speed $v_r$ of a $2,\text{kg}$ gun if it fires a $0.010,\text{kg}$ bullet at $400,\text{m/s}$ from rest.
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$v_r = -2,\text{m/s}$. $(2)(0) + (0.010)(0) = (2)v_r + (0.010)(400)$ gives $v_r = -2$ m/s.
$v_r = -2,\text{m/s}$. $(2)(0) + (0.010)(0) = (2)v_r + (0.010)(400)$ gives $v_r = -2$ m/s.
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Find the impulse $J$ on a $0.50,\text{kg}$ ball if it changes from $+4,\text{m/s}$ to $-2,\text{m/s}$.
Find the impulse $J$ on a $0.50,\text{kg}$ ball if it changes from $+4,\text{m/s}$ to $-2,\text{m/s}$.
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$J = -3,\text{N}\cdot\text{s}$. $J = (0.50)[(-2) - (+4)] = -3$ N·s.
$J = -3,\text{N}\cdot\text{s}$. $J = (0.50)[(-2) - (+4)] = -3$ N·s.
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What value of $e$ corresponds to a perfectly elastic collision?
What value of $e$ corresponds to a perfectly elastic collision?
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$e=1$. Objects separate with same relative speed as approach.
$e=1$. Objects separate with same relative speed as approach.
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What value of $e$ corresponds to a perfectly inelastic collision (objects stick together)?
What value of $e$ corresponds to a perfectly inelastic collision (objects stick together)?
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$e=0$. Objects stick together with no relative motion after collision.
$e=0$. Objects stick together with no relative motion after collision.
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Find $v$ if $m_1=3,\text{kg}$ at $u_1=2,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at $u_2=-2,\text{m/s}$.
Find $v$ if $m_1=3,\text{kg}$ at $u_1=2,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at $u_2=-2,\text{m/s}$.
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$v=0,\text{m/s}$. Equal and opposite momenta cancel when objects stick.
$v=0,\text{m/s}$. Equal and opposite momenta cancel when objects stick.
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Find the impulse on a $2,\text{kg}$ cart that changes velocity from $-1,\text{m/s}$ to $4,\text{m/s}$.
Find the impulse on a $2,\text{kg}$ cart that changes velocity from $-1,\text{m/s}$ to $4,\text{m/s}$.
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$J=10,\text{N}\cdot\text{s}$. $J=m\Delta v=(2)(4-(-1))=10$ N·s.
$J=10,\text{N}\cdot\text{s}$. $J=m\Delta v=(2)(4-(-1))=10$ N·s.
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Find $v$ if $m_1=1,\text{kg}$ at $u_1=8,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at $u_2=0$.
Find $v$ if $m_1=1,\text{kg}$ at $u_1=8,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at $u_2=0$.
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$v=2,\text{m/s}$. $(1)(8)+(3)(0)=(1+3)v$ gives $v=2$ m/s.
$v=2,\text{m/s}$. $(1)(8)+(3)(0)=(1+3)v$ gives $v=2$ m/s.
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Identify the correct 2D conservation equations if total initial momentum is only in $+x$ direction.
Identify the correct 2D conservation equations if total initial momentum is only in $+x$ direction.
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$p_{x,i}=p_{x,f}$ and $0=p_{y,f}$. $p_{y,i}=0$ so $p_{y,f}=0$; x-momentum conserved separately.
$p_{x,i}=p_{x,f}$ and $0=p_{y,f}$. $p_{y,i}=0$ so $p_{y,f}=0$; x-momentum conserved separately.
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What is the momentum of a particle of mass $m$ moving at velocity $v$ in 1D?
What is the momentum of a particle of mass $m$ moving at velocity $v$ in 1D?
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$p=mv$. Momentum equals mass times velocity in classical mechanics.
$p=mv$. Momentum equals mass times velocity in classical mechanics.
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What is the momentum-conservation equation for an isolated 1D collision of two objects?
What is the momentum-conservation equation for an isolated 1D collision of two objects?
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$m_1u_1+m_2u_2=m_1v_1+m_2v_2$. Total momentum before equals total momentum after for isolated systems.
$m_1u_1+m_2u_2=m_1v_1+m_2v_2$. Total momentum before equals total momentum after for isolated systems.
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What condition must be true for total momentum to be conserved during a collision?
What condition must be true for total momentum to be conserved during a collision?
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Net external impulse on the system is $0$. No external forces means momentum stays constant.
Net external impulse on the system is $0$. No external forces means momentum stays constant.
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What is the impulse-momentum theorem for one object in 1D?
What is the impulse-momentum theorem for one object in 1D?
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$J= riangle p=m(v-u)$. Impulse equals change in momentum.
$J= riangle p=m(v-u)$. Impulse equals change in momentum.
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What is the coefficient of restitution definition in 1D for two colliding objects?
What is the coefficient of restitution definition in 1D for two colliding objects?
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$e=\frac{v_2-v_1}{u_1-u_2}$. Measures relative separation speed over approach speed.
$e=\frac{v_2-v_1}{u_1-u_2}$. Measures relative separation speed over approach speed.
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State the formula for the common final velocity when two objects stick together in 1D.
State the formula for the common final velocity when two objects stick together in 1D.
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$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$. Total momentum divided by total mass gives common velocity.
$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$. Total momentum divided by total mass gives common velocity.
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Identify the sign rule for 1D momentum problems when choosing a positive direction.
Identify the sign rule for 1D momentum problems when choosing a positive direction.
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Velocities opposite the positive direction are negative. Ensures correct vector addition in 1D problems.
Velocities opposite the positive direction are negative. Ensures correct vector addition in 1D problems.
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