Velocity after collision (elasticity)

[DrMcDreamy]

This is a two part problem, but once I figure out one part I can do the other.

Homework Statement

a) A(n) 67.5 g object moving to the right at 26.7 cm/s overtakes and collides elastically with a(n) 18.8 g object moving in the same direction at 17.952 cm/s. Find the velocity of the first object immediately after the collision. Answer in units of cm/s.

b) Find the velocity of the second object immediately after the collision. Answer in units of cm/s.

Homework Equations

1) m₁(v_1i-v_1f) = m₂(v_2f-v_2i)
2) v_1i+v_1f = v_2f+v_2i

The Attempt at a Solution

1) (67.5 g)(26.7 cm/s - v_1f) = (18.8 g)(v_2f - 17.952 cm/s)
2) 26.7 cm/s + v_1f = v_2f +17.953 cm/s

(65.7g)(26.7cm/s) + 26.7cm/s = 18.8gv_2f
97.3 cm/s = v_2f

26.7 cm/s + v_1f = 97.3 cm/s + 17.952 cm/s
v_1f = 88.6 cm/s

Its wrong and I think my problem is somewhere in the math, I think I am messing up the calculations. Where am I wrong?

 

[alphysicist]

Hi DrMcDreamy,

This is a two part problem, but once I figure out one part I can do the other.

Homework Statement

a) A(n) 67.5 g object moving to the right at 26.7 cm/s overtakes and collides elastically with a(n) 18.8 g object moving in the same direction at 17.952 cm/s. Find the velocity of the first object immediately after the collision. Answer in units of cm/s.

b) Find the velocity of the second object immediately after the collision. Answer in units of cm/s.

Homework Equations

1) m₁(v_1i-v_1f) = m₂(v_2f-v_2i)
2) v_1i+v_1f = v_2f+v_2i

The Attempt at a Solution

1) (67.5 g)(26.7 cm/s - v_1f) = (18.8 g)(v_2f - 17.952 cm/s)
2) 26.7 cm/s + v_1f = v_2f +17.953 cm/s

(65.7g)(26.7cm/s) + 26.7cm/s = 18.8gv_2f

I don't believe this equation follows from the previous two equations. How did you get this one?

97.3 cm/s = v_2f

26.7 cm/s + v_1f = 97.3 cm/s + 17.952 cm/s
v_1f = 88.6 cm/s

Its wrong and I think my problem is somewhere in the math, I think I am messing up the calculations. Where am I wrong?

 

[spyrustheviru]

yes its the maths. try using leters while working on the equations, and calculate in the end.you'll see your mistake that way, and make your life easier

 

[DrMcDreamy]

So does the masses cancel out?

1) v1i-v1f = v2f-v2i
2) v1i+v1f = v2f+v2i

1) 26.7 cm/s - v_1f = v_2f - 17.952
2) 26.7 cm/s + v_1f = v_2f - 17.952

53.4 = 2v_2f
53.4/2 = v_2f
26.7 cm/s = v_2f

plug that into find v_1f:
26.7 cm/s + v_1f = 26.7 cm/s + 17.952
v_1f = 17.952 cm/s

Is that right? And if it is, qhy are the masses canceled out?

 

[DrMcDreamy]

^I put in the answers and it still came out wrong. Where am I going wrong?

 

[alphysicist]

So does the masses cancel out?

No the masses don't cancel. Your equations 1 and 2 from your original post:

1) (67.5 g)(26.7 cm/s - v1f) = (18.8 g)(v2f - 17.952 cm/s)
2) 26.7 cm/s + v1f = v2f +17.953 cm/s

looked right to me; the problem was that the next equation that you tried to get from these two was not correct.

My advice is to first divide both sides of eq. 1 by 67.5, giving:

26.7 cm/s - v1f = (18.8 /67.5 )(v2f - 17.952 cm/s)

and remember that the (18.8/67.5) multiplies both the v2f and the -17.952. Then eliminate v1f from this equation and equation 2. Does that help? What do you get?

1) v1i-v1f = v2f-v2i
2) v1i+v1f = v2f+v2i

1) 26.7 cm/s - v_1f = v_2f - 17.952
2) 26.7 cm/s + v_1f = v_2f - 17.952

53.4 = 2v_2f
53.4/2 = v_2f
26.7 cm/s = v_2f

plug that into find v_1f:
26.7 cm/s + v_1f = 26.7 cm/s + 17.952
v_1f = 17.952 cm/s

Is that right? And if it is, qhy are the masses canceled out?

 

[DrMcDreamy]

THANK YOU! I get it!

1) (67.5 g)(26.7 cm/s - v1f) = (18.8 g)(v2f - 17.952 cm/s)
2) 26.7 cm/s + v1f = v2f +17.952 cm/s

1) 26.7 cm/s - v1f = \frac{18.8g}{67.5g}(v2f - 17.952 cm/s)
26.7 cm/s - v1f = .279 v2f cm/s - 5.00 cm/s
2) 26.7 cm/s + v1f = v2f +17.952 cm/s

The v1f 's get canceled and end up with:

53.4 cm/s = 1.279 v2f cm/s + 12.952
-12.952 = 1.279 v2f

Divide by 1.279:

31.6 cm/s = v2f

And plug that into either equation to get v1f which is 22.9 cm/s

Both answers turn out to be correct! Thank you so much!

 

[alphysicist]

THANK YOU! I get it!

1) (67.5 g)(26.7 cm/s - v1f) = (18.8 g)(v2f - 17.952 cm/s)
2) 26.7 cm/s + v1f = v2f +17.952 cm/s

1) 26.7 cm/s - v1f = \frac{18.8g}{67.5g}(v2f - 17.952 cm/s)
26.7 cm/s - v1f = .279 v2f cm/s - 5.00 cm/s
2) 26.7 cm/s + v1f = v2f +17.952 cm/s

The v1f 's get canceled and end up with:

53.4 cm/s = 1.279 v2f cm/s + 12.952
-12.952 = 1.279 v2f

Divide by 1.279:

31.6 cm/s = v2f

And plug that into either equation to get v1f which is 22.9 cm/s

That all looks right to me.

Both answers turn out to be correct! Thank you so much!

Sure, glad to help!

 

[DrMcDreamy]

Ok, I have a similar problem and I did it the same way but the answer is wrong.

Problem:
A(n) 2.4 kg object moving at a speed of 4.4 m/s strikes a(n) 0.57 kg object initially at rest. Immediately after the collision, the 2.4 kg object has a velocity of 1.5 m/s directed 35◦ from its initial line of motion. What is the speed of the 0.57 kg object immediately after the collision? Answer in units of m/s.

Equations:

1) m1(v1i-v1f) = m2(v2f-v2i)
2) v1i+v1f = v2f+v2i

My Solution:

1) (2.4 kg)(4.4 m/s - 1.5 m/s) = (0.57 kg)(v_2f - 0 m/s)
2) 4.4 m/s + 1.5 m/s = v_2f +0 m/s

1) \frac{2.4 kg}{0.57 kg}(4.4 m/s - 1.5 m/s) = v_2f
4.2 (4.4 m/s - 1.5 m/s)= v_2f m/s
18.5 m/s - 6.3 m/s = v_2f
2) 4.4 m/s + 1.5 m/s = v_2f

18.1= 2v_2f
\frac{18.1}{2} = v_2f
9.05 = v_2f

I am wondering now if the " 35◦ from its initial line of motion" is taken into consideration since it is given, and if it does, how? TIA

 

[alphysicist]

Ok, I have a similar problem and I did it the same way but the answer is wrong.

Problem:
A(n) 2.4 kg object moving at a speed of 4.4 m/s strikes a(n) 0.57 kg object initially at rest. Immediately after the collision, the 2.4 kg object has a velocity of 1.5 m/s directed 35◦ from its initial line of motion. What is the speed of the 0.57 kg object immediately after the collision? Answer in units of m/s.

Equations:

1) m1(v1i-v1f) = m2(v2f-v2i)
2) v1i+v1f = v2f+v2i

These are the equations for a one-dimensional elastic collision. This problem has a two-dimensional collision, and I don't believe it is elastic. What is conserved in this problem? What equations do you use?

If this is not familiar, I would guess that your textbook probably has a section devoted to two-dimensional or glancing collisions.

My Solution:

1) (2.4 kg)(4.4 m/s - 1.5 m/s) = (0.57 kg)(v_2f - 0 m/s)
2) 4.4 m/s + 1.5 m/s = v_2f +0 m/s

1) \frac{2.4 kg}{0.57 kg}(4.4 m/s - 1.5 m/s) = v_2f
4.2 (4.4 m/s - 1.5 m/s)= v_2f m/s
18.5 m/s - 6.3 m/s = v_2f
2) 4.4 m/s + 1.5 m/s = v_2f

18.1= 2v_2f
\frac{18.1}{2} = v_2f
9.05 = v_2f

I am wondering now if the " 35◦ from its initial line of motion" is taken into consideration since it is given, and if it does, how? TIA

 

[DrMcDreamy]

These are the equations for a one-dimensional elastic collision. This problem has a two-dimensional collision, and I don't believe it is elastic. What is conserved in this problem? What equations do you use?

If this is not familiar, I would guess that your textbook probably has a section devoted to two-dimensional or glancing collisions.

Is momentum conserved? If it is, it would mean this problem is inelastic which means that when the masses collide, they stick to each other and I am assuming that they would have the same speed after the collision.

We didn't really do 2D collisions in lecture but I found an equation in my notes:

M₁V_1i + M₂V_2i = (M₁ + M₂)V

I also looked in my textbook and this is the equation I found:

P_AX + P_BX = P'_AX + P'_BX
p_x conserved: M_AV_A = M_AV'_Acos\theta'_A + M_BV'_Bcos\theta'_B

P_AY + P_BY = P'_AY + P'_BY
p_y conserved: 0 = M_AV'_Asin\theta'_A + M_BV'_Bsin\theta'_B

Problem:

A(n) 2.4 kg object moving at a speed of 4.4 m/s strikes a(n) 0.57 kg object initially at rest. Immediately after the collision, the 2.4 kg object has a velocity of 1.5 m/s directed 35◦ from its initial line of motion. What is the speed of the 0.57 kg object immediately after the collision? Answer in units of m/s.

Object 1:
m=2.4 kg; v_i=4.4 m/s; v_f=1.5 m/s
Object 2:
m=0.57 kg; v_i= 0, rest; v_f=?

My Solution:

I am confused, if this problem is inelastic, it means the masses are stuck together and would have the same speed after collision. It would mean V_2f = 1.5 m/s.

 

[alphysicist]

Is momentum conserved? If it is, it would mean this problem is inelastic which means that when the masses collide, they stick to each other

No, that's not correct. If momentum is conserved, but kinetic energy is not, then we call that inelastic.

Elastic collisions are when momentum and kinetic energy are conserved.

For the special case of when the objects stick together after the collision, then that is called "completely inelastic" or "perfectly inelastic" or some term like that.

So here the problem is inelastic, momentum is conserved, and the objects do not stick together.

and I am assuming that they would have the same speed after the collision.

We didn't really do 2D collisions in lecture but I found an equation in my notes:

M₁V_1i + M₂V_2i = (M₁ + M₂)V

Just to be clear, this equation is for the completely inelastic case, when the objects stick together after the collision. It does not apply to this problem.

I also looked in my textbook and this is the equation I found:

P_AX + P_BX = P'_AX + P'_BX
p_x conserved: M_AV_A = M_AV'_Acos\theta'_A + M_BV'_Bcos\theta'_B

P_AY + P_BY = P'_AY + P'_BY
p_y conserved: 0 = M_AV'_Asin\theta'_A + M_BV'_Bsin\theta'_B

Yes, these are the equations that apply to this problem.

These are the general equations for two-dimensional momentum conservation, in the x and y directions:

P_AX + P_BX = P'_AX + P'_BX
P_AY + P_BY = P'_AY + P'_BY

and your other two equations are what you get when you apply those two equations to this problem:

p_x conserved: M_AV_A = M_AV'_Acos\theta'_A + M_BV'_Bcos\theta'_B
p_y conserved: 0 = M_AV'_Asin\theta'_A + M_BV'_Bsin\theta'_B

If you apply these equations to your problem, you should get the answer. (With these equations written this way, you might need to be careful with the unknown angle for the second object.)

Problem:

A(n) 2.4 kg object moving at a speed of 4.4 m/s strikes a(n) 0.57 kg object initially at rest. Immediately after the collision, the 2.4 kg object has a velocity of 1.5 m/s directed 35◦ from its initial line of motion. What is the speed of the 0.57 kg object immediately after the collision? Answer in units of m/s.

Object 1:
m=2.4 kg; v_i=4.4 m/s; v_f=1.5 m/s
Object 2:
m=0.57 kg; v_i= 0, rest; v_f=?

My Solution:

I am confused, if this problem is inelastic, it means the masses are stuck together and would have the same speed after collision. It would mean V_2f = 1.5 m/s.

 

[DrMcDreamy]

If you apply these equations to your problem, you should get the answer. (With these equations written this way, you might need to be careful with the unknown angle for the second object.)

Would the unknown angle be 0◦ ? Since both masses are in the same plane at first and than the first object collides with the second, which is at rest. The first object is directed 35◦ from the initial line of motion. Would the second object still be in the same line of motion?

 

[alphysicist]

Would the unknown angle be 0◦ ? Since both masses are in the same plane at first and than the first object collides with the second, which is at rest. The first object is directed 35◦ from the initial line of motion. Would the second object still be in the same line of motion?

Your y-equation tells you that that is impossible. If you plug in zero degrees for the second term on the right, that term becomes zero; but the y-equations says that the second term must be the negative of the first (or else they will not equal the left hand side, which is zero).

Physically speaking: before the collision there was no momentum in the y-direction. Since the momentum in the y-direction in this problem is conserved, the total y-momentum after the collision must also be zero. So if the first object ends up moving in the positive y-direction after the collision, there must be a compensating motion in the negative y-direction for the second object, or else momentum the momentum would not be conserved.

So the unknown angle is not zero degrees; you need to find out what angle it is. However, in your two equations, you only have two unknowns, so you can go ahead and solve for them as an algebra problem.

 

[DrMcDreamy]

Ok, I figured it out! My professor went over it in class and here's my work:

Sorry if you can't read my handwriting!

 

[alphysicist]

Ok, I figured it out! My professor went over it in class and here's my work:

Sorry if you can't read my handwriting!

Sorry, I've been away for a couple of days. But your work looks right to me!