Problem #1
A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.030 kg shooter marble moving to the left at 0.180 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0.030 kg marble after the collision?Answer:
Given: m1 = 0.015 kg, m2 = 0.030 kg, v1,i = 0.225 m/s to the right, v1,i = +0.225 m/s, v2,i = 0.180 m/s to the left, v2,i = −0.180 m/s, v1,f = 0.315 m/s to the left, v1,f = −0.315 m/s, as shown in Fig. 1
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| Fig.1 |
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Rearrange the equation to isolate the final velocity of m2,
m2v2,f = m1v1,i + m2v2,i − m1v1,f
v2,f = (m1v1,i + m2v2,i − m1v1,f)/m2
Substitute the values into the equation and solve: The rearranged conservation-of-momentum equation will allow you to isolate and solve for the final velocity.
v2,f = [(0.015 kg)(0.225 m/s) + (0.030 kg)(−0.180 m/s) − (0.015 kg)(−0.315 m/s)]/0.030 kg
v2,f = 9.0 × 10−2 m/s to the right
Problem #2
Consider a frictionless track ABC as shown in Fig. 2. A block of mass m1 = 5.001 kg is released from A. It makes a head–on elastic collision with a block of mass m2 = 10.0kg at B, initially at rest. Calculate the maximum height to which m1 rises after the collision.
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| Fig.2 |
Whoa! What is this problem talking about?? We release mass m1; it slides down to the slope, picking up speed, until it reaches B. At B it makes a collision with mass m2, and we are told it is an elastic collison. The last sentence in the problem implies that in this collision m1 will reverse its direction of motion and head back up the slope to some maximum height. We would also guess that m2 will be given a forward velocity.
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| Fig.3 |
This sequence is shown in Fig. 3. First we think about the instant of time just before the collision. Mass m1 has velocity v1i and mass m2 is still stationary. How can we find v1i?
We can use the fact that energy is conserved as m1 slides down the smooth (frictionless) slope. At the top of the slope m1 had some potential energy, U = m1gh (with h = 5.00 m) which is changed to kinetic energy,
K = ½ m1v21i
when it reaches the bottom. Conservation of energy gives us:
m1gh = ½ mv21i
v21i = 2gh = 2(9.80 m/s2)(5.00 m) = 98.0 m2/s2
so that
v1i = +9.90 m/s
We chose the positive value here since m1 is obviously moving forward at the bottom of the
slope. So m1’s velocity just before striking m2 is +9.90 m/s .
Now m1 makes an elastic (one–dimensional) collision with m2. What are the final velocities of the masses? For this we can use the result , using v2i = 0. We get:
v1f = (m1 − m2)v1i/(m1 + m2)
= (5.001 kg − 10.0kg)(+9.90 m/s)/(5.001 kg + 10.0kg)
v1f = −3.30 m/s
and
v2f = (2m1)(v1i)/(m1 + m2)
= 2(5.001 kg) (+9.90 m/s)/(5.001 kg + 10.0kg)
v2f = +6.60 m/s
So after the collision, m1 has a velocity of −3.30 m/s ; that is, it has speed 3.30 m/s and it is now moving to the left . After the collision, m2 has velocity +6.60 m/s , so that it is moving to the right with speed 6.60 m/s .
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| Fig.4 |
Since m1 is now moving to the left, it will head back up the slope. (See Fig. 4.) How high will it go? Once again, we can use energy conservation to give us the answer. For the trip back up the slope, the initial energy (all kinetic) is
Ei = Ki = ½ m(3.30 m/s)2
and when it reaches maximum height (h) its speed is zero, so its energy is the potential energy,
Ef = Uf = mgh
Conservation of energy, Ei = Ef gives us:
½ m(3.30 m/s)2 = mgh
h = (3.30 m/s)2/2g = 0.556 m
Mass m1 will travel back up the slope to a height of 0.556 m.
Problem #3
A 5.0kg block slides from rest down an L = 2.50 m long 25° incline, Fig.5. At the bottom it undergoes an elastic collision with a 10.0kg block sending it towards a 35° incline. After the collision, how far along its incline does each block go? The surface is frictionless.
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| Fig.5 |
We have a change in height and speed in the first part of the problem, so that suggests that we have a WorkEnergy problem. In the second part of the problem, there is a collision which suggest that we use conservation of momentum. In the final portion, there is a change in height and speed again, so this suggests that we have a WorkEnergy problem.
(i) Since there is no friction, WNC = 0. Hence Ef = Ei or
½m1v2 = m1gh
The height h is related to L by h = Lsin(25°). Substituting in this relation, and rearranging to get v by itself yields,
v = [2gLsin(25°)]1/2 = [2(9.81)(2.5) sin(25°)]1/2 = 4.553 m/s
This is the velocity of the first block just before the collision. This velocity will be the initial velocity for part (ii).
(ii) In an elastic collision, both momentum and kinetic energy is conserved. Thus we have the equations;
m1v1f + m2v2f = m1v1i + m2v2i , (1)
v1f – v2f = –(v1i – v2i) (2)
Since v2i = 0, the two equations can be combine to yield
v1f = (m1 – m2)v1i/(m1 + m2)
v1f = (5 kg – 10 kg)(4.553 m/s)/(5 kg + 10 kg) = –1.518 m/s
and
v2f = (2m2)v1i/(m1 + m2) = (2 x 5 kg)(4.553 m/s)/(5 kg + 10 kg) = 3.035 m/s
So the first block will bounce backwards and return up the incline it came down. The second block will move up the incline on the right
(iii) Applying conservation of energy for the first block
½m1v2 = m1gL1 sin(25°)
Solving for L1 , we find
L1 = v2/(2gsin(25°)) = (1.518)2/(29.81 sin(25°)) = 0.278 m
Similarly, the second block moves up its incline a distance
L2 = v2/(2gsin(35°)) = (3.035)2/(29.81 sin (35°)) = 0.819 m
So the first block moves 0.28 m up the left incline after the collision while the second block moves 0.82 m up the right incline.
Problem #4
A ball 1 depends on the small stem 1 m in length. Then this ball is pounded by ball 2 whose mass is twice the mass of the ball 1. Hit the speed of ball 2 so that ball 1 reaches the highest point of the circular path. Suppose the collision is perfect. (Rod mass ignored)
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| Fig.6 |
Logic: In order for the ball to reach the highest point of the trajectory, the speed at the highest point is at least zero, vB = 0. With the energy-effort formula we can calculate the speed at A, vA. This vA speed is the speed after the collision. To calculate the velocity before the collision we use the restitution formula and the conservation of momentum.
Our analysis is divided into 2 parts: Ball movements from A to B and collisions in A.
Movement from A to B
The effort-energy formula gives
Wgrav = ΔK
–ΔP = ΔK
–mg(hB – hA) = ½mvB2 – ½ mvA2
–g(hB – hA) = – ½vA2
vA = [2g(hB - hA)]1/2 = [2 x 9.8 m/s2(2 m – 0)]1/2 = 6.32 m/s
Ball up if vA ≥ 6.32 m/s
Collision:
Given: m1 = m, m2 = 2m, v1f = vA = 6.32 m/s and v1i = 0.
Conservation of momentum
m1v1i + m2v2i = m1v1f + m2v2f
m . 0 + 2mv2i = m . 6.32 m/s + 2mv2f
3.16 = v2i – v2f (1)
coefficient of restitution, e = 1,
e = -(v2f – v1f)/(v2i – v1i) = 1
(v2i – v1i) = -(v2f – v1f)
(v2i – 0) = -(v2f – 6.32)
6.32 = v2i + v2f (2)
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| Fig.7 |
3.16 = v2i – v2f
6.32 = v2i + v2f
2v2i = 9.48
v2i = 4.74 m/s
So, the speed of the object pounding is 4.71 m / s
Note: if the rod is replaced by a rope, the condition of the object reaching the highest point is T = 0
Problem #5
Both dropped at same time from height h (have a small gap and h >> Diameter of balls). Large ball rebounds elastically from floor, then the small ball rebounds elastically from large ball. a)What m would cause the large ball to stop when the two collide? b) What height does the small ball then reach?. (Parameters: M=0.63kg, h=1.8m)
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| Fig.8 |
Step 1: Get velocities of both balls when they hit from energy conservation Kf = Pi
0.5Mv2= Mgh
v = (2gh)1/2 = (2 x 9.8 m/s2 x 1.8 m)1/2 = 5.94 m/s
independent of mass: v = –5.94m/s [(–) because it goes down]
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| Fig.9 |
Collision of large ball with larger object (Earth): Ball turns around, same speed,
vMi = 5.94 m/s and vmi = –5.94m/s
Step 3: Elastic collision betw. large and small ball. Momentum and energy conservation should allow to solve this.
Elastic collision betw. large and small ball. Use:
vM,f = (M – m)vMi/(M + m) + 2mvmi/(M + m)
Solve: vMf = 0 with: vMi =- vmi leads to m = M/3 = 0.21 kg
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| Fig.10 |
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| Fig.11 |
vM,f = (m – M)vmi/(M + m) + 2MvMi/(M + m)
Use vMi = - vmi and M = 3m leads to vmf = 2vMi = 11.88 m/s
K = ½ mv2mf = mgH = U
H = v2mf/2g = 4h = 7.2 m
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