1983 Exam Question w/SHM and conservation of momentum

So by equating the two, you can solve for x. In summary, a block of mass M collides with a second block of mass 2M and sticks to it on a horizontal, frictionless table attached to a relaxed spring of spring constant k. To find the speed of the blocks immediately after impact (v), use conservation of momentum to get 2/3vo. For the maximum distance the spring is compressed (x), equate the initial kinetic energy of the blocks to the potential energy of the spring (.5mv2 = .5kx2) and solve for x. And finally, to find the period of the subsequent simple harmonic motion (T), use the equation T = 2pi*sqrt(m/k
  • #1
rvhockey
11
0
A block of mass M is resting on a horizontal, frictionless table and is attached to a relaxed spring of spring constant k. A second block of mass 2M and initial speed vo collides with and sticks to the first block. Develop expressions for the following quantities in terms of M, k, and Vo.

a) v, the speed of the blocks immediately after impact
b) x, the maximum distance the spring is compressed
c) T, the period of the subsequent simple harmonic motion




m1v1+m2v2=m1v1'+m2v2'
EPE=.5kx2
KE = .5mv2
T = 2pi*sqrt(m/k)




I'm only having trouble with part b. I knew to use conservation of momentum for a and get 2/3vo. c was easy as you could just use the last equation and plug in 3M and k. But for b I'm not sure if I'm doing it right. What I'm doing it plugging in 3M and 2/3vo to solve for the initial KE and setting it equal to .5kx2 to get x. Is this right?
 
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  • #2
Looks right to me. All that's happening is the kinetic energy of the blocks is transforming into the potential energy of the spring.
 
  • #3


Yes, you are on the right track. To solve for the maximum distance the spring is compressed, you need to use the conservation of energy principle. Initially, the system has only kinetic energy (KE) due to the initial speed of the second block. After the collision, the system has both kinetic and potential energy (PE) due to the compressed spring. Therefore, the initial KE must be equal to the final KE and PE. This can be represented by the equation:

.5m1v1^2 + .5m2v2^2 = .5kx^2

Substituting in the values of m1, m2, and v2 (which is equal to 2/3vo), we get:

.5(M)(0)^2 + .5(2M)(2/3vo)^2 = .5kx^2

Solving for x, we get:

x = (2/3vo)^2 / k

Therefore, the maximum distance the spring is compressed is dependent on the mass of the block and the initial speed, but is inversely proportional to the spring constant.
 

1. What is SHM?

SHM stands for Simple Harmonic Motion. It is a type of periodic motion in which an object oscillates back and forth around an equilibrium point. It is characterized by a sinusoidal displacement, velocity, and acceleration.

2. How is SHM related to the 1983 Exam Question?

The 1983 Exam Question involves a scenario in which a mass is attached to a spring and undergoes SHM. The question asks for the period and amplitude of the motion.

3. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant. This means that in a system where there is no external force acting on the objects, the initial momentum will be equal to the final momentum.

4. How is conservation of momentum applied to the 1983 Exam Question?

In the 1983 Exam Question, conservation of momentum is used to calculate the velocity of the mass after it collides with another mass. The total momentum before the collision is equal to the total momentum after the collision.

5. What are the key equations used in solving the 1983 Exam Question?

The key equations used in solving the 1983 Exam Question are the equation for SHM (x = A sin(ωt)), and the equation for conservation of momentum (m1v1i + m2v2i = m1v1f + m2v2f). Other equations related to SHM and conservation of momentum may also be used depending on the specific scenario presented in the question.

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