Question

(Elastic collision) Particle 1 of mass $m_1$ and kinetic energy $K_{1,i}$ collides elastically with stationary particle 2 of mass $m_2$. After the collision, particle 2 moves under angle $phi$ with respect to the original direction of the motion of particle 1. (a) Calculate $v_{2,f}$ as a function of $v_{1,i}$ and angle $phi$. Then show that the kinetic energy $K_{2,f}$ delivered to particle 2 is $K_{2,f} = K_{1,i} frac{4m_1m_2}{(m_1 + m_2)^2} cos^2(phi)$. Hint: Square both equations for momentum conservation and add them together. (b) Show that $K_{2,f}$ has the maximum value in a head-on collision. (c) Show that in a head-on collision, the energy lost by particle 1 is given by $K_{1,i} - K_{1,f} = frac{4m_1m_2}{(m_1 + m_2)^2} K_{1,i}$.

          (Elastic collision)
Particle 1 of mass $m_1$ and kinetic energy $K_{1,i}$ collides elastically with stationary particle 2
of mass $m_2$. After the collision, particle 2 moves under angle $phi$ with respect to the original
direction of the motion of particle 1.
(a) Calculate $v_{2,f}$ as a function of $v_{1,i}$ and angle $phi$. Then show that the kinetic energy $K_{2,f}$
delivered to particle 2 is
$K_{2,f} = K_{1,i} frac{4m_1m_2}{(m_1 + m_2)^2} cos^2(phi)$.
Hint: Square both equations for momentum conservation and add them together.
(b) Show that $K_{2,f}$ has the maximum value in a head-on collision.
(c) Show that in a head-on collision, the energy lost by particle 1 is given by
$K_{1,i} - K_{1,f} = frac{4m_1m_2}{(m_1 + m_2)^2} K_{1,i}$.

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$(Elastic collision) Particle 1 of mass m1 and kinetic energy K1,i collides elastically with stationary particle 2 of mass m2. After the collision, particle 2 moves under angle phi with respect to the original direction of the motion of particle 1. (a) Calculate v2,f as a function of v1,i and angle phi. Then show that the kinetic energy K2,f delivered to particle 2 is K2,f = K1,i frac4m1m2(m1 + m2)^2 cos^2(phi). Hint: Square both equations for momentum conservation and add them together. (b) Show that K2,f has the maximum value in a head-on collision. (c) Show that in a head-on collision, the energy lost by particle 1 is given by K1,i - K1,f = frac4m1m2(m1 + m2)^2 K1,i.$

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University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

08/01/2023

Step 1

First, we need to use conservation of momentum to find the final velocities of both particles. Since the collision is elastic, we also know that kinetic energy is conserved. Conservation of momentum: m1V1,i = m1V1,f cos(θ) + m2V2,f Conservation of kinetic Show more…

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(Elastic collision) Particle 1 of mass m1 and kinetic energy K1,i collides elastically with stationary particle 2 of mass m2. After the collision, particle 2 moves under angle ϕ with respect to the original direction of the motion of particle 1. (a) Calculate v2,f as a function of v1,i and angle ϕ. Then show that the kinetic energy K2,f delivered to particle 2 is K2,f = K1,i * (4m1m2 / (m1 + m2)^2) * cos^2(ϕ). Hint: Square both equations for momentum conservation and add them together. (b) Show that K2,f has the maximum value in a head-on collision. (c) Show that in a head-on collision, the energy lost by particle 1 is given by K1,i - K1,f = (4m1m2 / (m1 + m2)^2) * K1,i.

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Transcript

00:01 So, here we are given the particle 1 whose mass is m1 and the kinetic energy is k1 of i which collide elastically with the particle 2.

00:13 So, it collide with the particle 2 after whose mass is m2 and collide with the particle which is making an angle phi.

00:24 So, we are considering about the part 1 where we are considering for v2 and f as a function of v and phi.

00:30 Then we have to show that the kinetic energy delivered to the particle 2 is represented as k2f that is equal to k1 of i 4 of m1 m2 which is divided by m1 plus m2 to its whole square cos square of phi.

00:46 So, this is given to us.

00:48 So, here this value is given.

00:49 We are considering about the two particles from here.

00:52 So, two particles here can be given as that we are considering the energy conservation.

00:59 So, k1 of i is given as 1 divided by 2 m1 v of i 1 to its whole square and the angle is same to its phi.

01:09 So, we are having two particles.

01:11 So, m1 v1 of i plus m2 v2 of i from here is equals to m1 v1 of f plus m2 v2 of f multiplied by the cos of phi.

01:23 Now, we have to multiply both by the side 1 divided by 2 and then squaring and adding.

01:27 So, this from here is equals to 1 divided by 4 m1 v1 of i square that is less than equals to 1 divided by 4 of m1 v1 of f plus m2 v2 of f multiplied by the cos of phi to its whole square.

01:42 Solving the term from here, we get the value that is equals to m1 k1 of k of i1 that is equals to 1 divided by 4 m2 k of f2 cos square of phi plus m2 multiplied by the v2 of f multiplied by the cos of phi.

01:59 Again solving the term from here, this is equals to vm1 k i1 that is equals to m2 divided by the cos of phi multiplied by the k of f2 plus 2 cos of phi multiplied by the v2 of f that is divided by the cos of phi.

02:17 So, solving the term from here, we get the value of k2 of f that is equals to k1 k of i1 multiplied by the cos square of phi which is multiplied by the 4 of m1 m2 which is divided by the m1 plus m2 to its whole square.

02:32 So, this is the answer to the part of the question.

02:35 Now, we are considering about the part b.

02:38 In the part b, we have to show that k2 of f has the maximum value in the head -on collision...

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