Perfectly elastic collision

In summary, the problem involves a head-on, perfectly elastic collision between a 15 g ball and a 117 g ball hanging motionless from a 1.53-m-long string. After the collision, the 117 g ball swings out to a maximum angle of 53°. To find the initial speed of the 15 g ball, we can use the equations for conservation of energy and momentum, taking into consideration the height of the hanging ball as well. This results in 2 equations and 2 unknowns, one of which is the initial speed Vo that is being asked for.
  • #1
sktgurl930
21
0

Homework Statement


A 15 g ball is fired horizontally with speed v0 toward a 117 g ball hanging motionless from a 1.53-m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 117 g ball swings out to a maximum angle θmax=53°. What was v0?

Homework Equations


h=L(L-cos(theta)
V=Square root of (m*g*h/(.5*M) little m is ball moving and M is the ball not moving
(v2x)f=2m1/m1+m2 (V1x)i



The Attempt at a Solution


Im not sure if I am using the right equations and whether the angle given is the one we use in the equation, or if we have to subtract it from 90 first
 
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  • #2
sktgurl930 said:

Homework Statement


A 15 g ball is fired horizontally with speed v0 toward a 117 g ball hanging motionless from a 1.53-m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 117 g ball swings out to a maximum angle θmax=53°. What was v0?

2. Homework Equations
h=L(L-cos(theta)
V=Square root of (m*g*h/(.5*M) little m is ball moving and M is the ball not moving
(v2x)f=2m1/m1+m2 (V1x)i

The Attempt at a Solution


Im not sure if I am using the right equations and whether the angle given is the one we use in the equation, or if we have to subtract it from 90 first

A couple of things.

Initial KE is 1/2*m*v2

The second is that the angle because it is hanging is with the vertical. Hence the height should be given by the Cosθ times the string length subtracted from the length. (h = length - projection of the string to the vertical.)

So ... 1/2*m1*v2 = m2*g*(L - L*cos53)

v2 = 2*(m2/m1)*g*(L-L*Cos53)
 
  • #3
LowlyPion said:
A couple of things.

Initial KE is 1/2*m*v2

The second is that the angle because it is hanging is with the vertical. Hence the height should be given by the Cosθ times the string length subtracted from the length. (h = length - projection of the string to the vertical.)

So ... 1/2*m1*v2 = m2*g*(L - L*cos53)

v2 = 2*(m2/m1)*g*(L-L*Cos53)

so my mass 1 is the ball not moving right??
is this the equation i use to get the answer or do i have to plug it into another one
 
  • #4
sktgurl930 said:
so my mass 1 is the ball not moving right??
is this the equation i use to get the answer or do i have to plug it into another one

Sorry my mistake. The m1 I wrote on the left should be m2 - the stationary ball. We are trying to calculate the velocity of m2 AFTER impact. The m's should cancel there. Sorry for my typo which I then proceeded to run with.

With that final velocity for the stationary ball after impact and the usual equations for conservation of energy and momentum, you should now have 2 equations and 2 unknowns, one of which is the Vo that they ask for.
 

1. What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision between two objects where the total kinetic energy of the system is conserved. This means that no energy is lost or gained during the collision, and the objects bounce off each other without any deformation.

2. What is the difference between a perfectly elastic collision and an inelastic collision?

In a perfectly elastic collision, the total kinetic energy of the system is conserved. In an inelastic collision, some kinetic energy is lost due to deformation or other factors, resulting in the objects sticking together after the collision.

3. Are perfectly elastic collisions common in real life?

No, perfectly elastic collisions are rare in real life. In most collisions, some energy is lost due to factors such as friction, air resistance, and deformation of the objects involved.

4. How do you calculate the velocities of objects after a perfectly elastic collision?

The velocities of the objects after a perfectly elastic collision can be calculated using the conservation of momentum and the conservation of kinetic energy equations. These equations take into account the masses and velocities of the objects before and after the collision.

5. Can an object have a perfectly elastic collision with itself?

No, an object cannot have a perfectly elastic collision with itself. In order for a collision to be considered perfectly elastic, there must be two objects involved that exchange energy during the collision.

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