Determining final speed of a billiard ball after elastic collision
up vote
1
down vote
favorite
A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?
I set up two relationships involving momentum as follows
Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector
Based on conservation of momentum:
$$1) quad(3m/s)m=mv_1cos(30^circ)+mv_2cos(theta)$$
$$(3m/s)=v_1cos(30^circ)+v_2cos(theta)$$
$$2) quad v_1sin(30^circ)=v_2sin(theta)$$
However this is as far as I got and I don't know how to solve for $v_2$.
Am I missing something?
Btw I know that the answer is 1.5m/s.
classical-mechanics
edited Nov 13 at 1:02
David G. Stork
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
add a comment |
up vote
1
down vote
favorite
A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?
I set up two relationships involving momentum as follows
Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector
Based on conservation of momentum:
$$1) quad(3m/s)m=mv_1cos(30^circ)+mv_2cos(theta)$$
$$(3m/s)=v_1cos(30^circ)+v_2cos(theta)$$
$$2) quad v_1sin(30^circ)=v_2sin(theta)$$
However this is as far as I got and I don't know how to solve for $v_2$.
Am I missing something?
Btw I know that the answer is 1.5m/s.
classical-mechanics
edited Nov 13 at 1:02
David G. Stork
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?
I set up two relationships involving momentum as follows
Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector
Based on conservation of momentum:
$$1) quad(3m/s)m=mv_1cos(30^circ)+mv_2cos(theta)$$
$$(3m/s)=v_1cos(30^circ)+v_2cos(theta)$$
$$2) quad v_1sin(30^circ)=v_2sin(theta)$$
However this is as far as I got and I don't know how to solve for $v_2$.
Am I missing something?
Btw I know that the answer is 1.5m/s.
classical-mechanics
edited Nov 13 at 1:02
David G. Stork
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?
I set up two relationships involving momentum as follows
Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector
Based on conservation of momentum:
$$1) quad(3m/s)m=mv_1cos(30^circ)+mv_2cos(theta)$$
$$(3m/s)=v_1cos(30^circ)+v_2cos(theta)$$
$$2) quad v_1sin(30^circ)=v_2sin(theta)$$
However this is as far as I got and I don't know how to solve for $v_2$.
Am I missing something?
Btw I know that the answer is 1.5m/s.
classical-mechanics
classical-mechanics
edited Nov 13 at 1:02
David G. Stork
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
edited Nov 13 at 1:02
David G. Stork
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
edited Nov 13 at 1:02
David G. Stork
8,97621232
edited Nov 13 at 1:02
David G. Stork
8,97621232
edited Nov 13 at 1:02
David G. Stork
8,97621232
8,97621232
asked Nov 13 at 0:53
Anson Pang
523
asked Nov 13 at 0:53
Anson Pang
523
asked Nov 13 at 0:53
Anson Pang
523
523
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13
add a comment |
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.
The three unknowns are $v_1$, $v_2$ and $theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.
answered Nov 13 at 1:02
tom
1429
add a comment |
up vote
0
down vote
Momentum and energy must be conserved.
$v_1 cos 30^circ + v_2 cos theta = 3\
v_1 sin 30^circ + v_2 sin theta = 0\
v_1^2 + v_2^2 = 9$
The first two equations describe the momentum of the system, the last describes the energy.
$theta = arcsin frac {v_1}{2v_2}\
v_1 frac {sqrt 3}{2} + v_2 sqrt {1-frac {v_1^2}{4v_2^2}} = 3\
v_1 sqrt 3+ sqrt {4v_2^2-v_1^2} = 6\
sqrt {36 - 5v_1^2} = 6 - v_1sqrt 3\
36 - 5v_1^2 = 36 - v_112sqrt 3 + 3v_1^2\
12sqrt 3 v_1 = 8v_1^2\
v_1 = 0 text { or } v_1 = frac {3sqrt 3}{2}$
If $v_1 = 0$ then it isn't rolling away at a $30^circ$ deflection.
$v_2 = 9 - frac {27}{4} = 2.25$
answered Nov 13 at 1:37
Doug M
42.6k31752
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.
The three unknowns are $v_1$, $v_2$ and $theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.
answered Nov 13 at 1:02
tom
1429
add a comment |
up vote
0
down vote
If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.
The three unknowns are $v_1$, $v_2$ and $theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.
answered Nov 13 at 1:02
tom
1429
add a comment |
up vote
0
down vote
up vote
0
down vote
If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.
The three unknowns are $v_1$, $v_2$ and $theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.
answered Nov 13 at 1:02
tom
1429
If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.
The three unknowns are $v_1$, $v_2$ and $theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.
answered Nov 13 at 1:02
tom
1429
answered Nov 13 at 1:02
tom
1429
answered Nov 13 at 1:02
tom
1429
answered Nov 13 at 1:02
tom
1429
1429
add a comment |
add a comment |
up vote
0
down vote
Momentum and energy must be conserved.
$v_1 cos 30^circ + v_2 cos theta = 3\
v_1 sin 30^circ + v_2 sin theta = 0\
v_1^2 + v_2^2 = 9$
The first two equations describe the momentum of the system, the last describes the energy.
$theta = arcsin frac {v_1}{2v_2}\
v_1 frac {sqrt 3}{2} + v_2 sqrt {1-frac {v_1^2}{4v_2^2}} = 3\
v_1 sqrt 3+ sqrt {4v_2^2-v_1^2} = 6\
sqrt {36 - 5v_1^2} = 6 - v_1sqrt 3\
36 - 5v_1^2 = 36 - v_112sqrt 3 + 3v_1^2\
12sqrt 3 v_1 = 8v_1^2\
v_1 = 0 text { or } v_1 = frac {3sqrt 3}{2}$
If $v_1 = 0$ then it isn't rolling away at a $30^circ$ deflection.
$v_2 = 9 - frac {27}{4} = 2.25$
answered Nov 13 at 1:37
Doug M
42.6k31752
add a comment |
up vote
0
down vote
Momentum and energy must be conserved.
$v_1 cos 30^circ + v_2 cos theta = 3\
v_1 sin 30^circ + v_2 sin theta = 0\
v_1^2 + v_2^2 = 9$
The first two equations describe the momentum of the system, the last describes the energy.
$theta = arcsin frac {v_1}{2v_2}\
v_1 frac {sqrt 3}{2} + v_2 sqrt {1-frac {v_1^2}{4v_2^2}} = 3\
v_1 sqrt 3+ sqrt {4v_2^2-v_1^2} = 6\
sqrt {36 - 5v_1^2} = 6 - v_1sqrt 3\
36 - 5v_1^2 = 36 - v_112sqrt 3 + 3v_1^2\
12sqrt 3 v_1 = 8v_1^2\
v_1 = 0 text { or } v_1 = frac {3sqrt 3}{2}$
If $v_1 = 0$ then it isn't rolling away at a $30^circ$ deflection.
$v_2 = 9 - frac {27}{4} = 2.25$
answered Nov 13 at 1:37
Doug M
42.6k31752
add a comment |
up vote
0
down vote
up vote
0
down vote
Momentum and energy must be conserved.
$v_1 cos 30^circ + v_2 cos theta = 3\
v_1 sin 30^circ + v_2 sin theta = 0\
v_1^2 + v_2^2 = 9$
The first two equations describe the momentum of the system, the last describes the energy.
$theta = arcsin frac {v_1}{2v_2}\
v_1 frac {sqrt 3}{2} + v_2 sqrt {1-frac {v_1^2}{4v_2^2}} = 3\
v_1 sqrt 3+ sqrt {4v_2^2-v_1^2} = 6\
sqrt {36 - 5v_1^2} = 6 - v_1sqrt 3\
36 - 5v_1^2 = 36 - v_112sqrt 3 + 3v_1^2\
12sqrt 3 v_1 = 8v_1^2\
v_1 = 0 text { or } v_1 = frac {3sqrt 3}{2}$
If $v_1 = 0$ then it isn't rolling away at a $30^circ$ deflection.
$v_2 = 9 - frac {27}{4} = 2.25$
answered Nov 13 at 1:37
Doug M
42.6k31752
Momentum and energy must be conserved.
$v_1 cos 30^circ + v_2 cos theta = 3\
v_1 sin 30^circ + v_2 sin theta = 0\
v_1^2 + v_2^2 = 9$
The first two equations describe the momentum of the system, the last describes the energy.
$theta = arcsin frac {v_1}{2v_2}\
v_1 frac {sqrt 3}{2} + v_2 sqrt {1-frac {v_1^2}{4v_2^2}} = 3\
v_1 sqrt 3+ sqrt {4v_2^2-v_1^2} = 6\
sqrt {36 - 5v_1^2} = 6 - v_1sqrt 3\
36 - 5v_1^2 = 36 - v_112sqrt 3 + 3v_1^2\
12sqrt 3 v_1 = 8v_1^2\
v_1 = 0 text { or } v_1 = frac {3sqrt 3}{2}$
If $v_1 = 0$ then it isn't rolling away at a $30^circ$ deflection.
$v_2 = 9 - frac {27}{4} = 2.25$
answered Nov 13 at 1:37
Doug M
42.6k31752
answered Nov 13 at 1:37
Doug M
42.6k31752
answered Nov 13 at 1:37
Doug M
42.6k31752
answered Nov 13 at 1:37
Doug M
42.6k31752
42.6k31752
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2996130%2fdetermining-final-speed-of-a-billiard-ball-after-elastic-collision%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
you are missing that energy is conserved in an elastic collision and $v_1^2 + v_2^2 = 3^2$
– Doug M
Nov 13 at 1:13