Elastic Collision: Mass m2 Velocity & Momentum of m1

AI Thread Summary
In an elastic collision where mass m1 approaches a stationary mass m2 (with m2 > m1), both momentum and kinetic energy are conserved. The final velocity of m2 will be lower than the initial velocity of m1, as momentum must remain constant. The momentum of m1 after the collision can be calculated using conservation equations, indicating that it will be greater than its initial momentum due to the positive value of the momentum transfer from m2. Mathematically, this is supported by the conservation of kinetic energy, which shows that the final velocity of m1 will exceed its initial velocity. Overall, the collision results in m1 gaining momentum while m2 moves with a reduced velocity.
chynawok
Messages
4
Reaction score
0
Suppose that a particle of mass m1 approaches a stationary mass m2 and that m2 > m1.
a) Describe the velocity of m2 after an elastic collision--that is, one in which both momentum and kinetic energy are conserved. Justify the answer mathematically.
b) What is the approximate momentum of m1 after the collision?
 
Physics news on Phys.org
Read the rules, you need to show some work, start setting up the equations of conservation.
 


a) In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision are equal. In this scenario, the initial momentum of m1 is given by its mass (m1) multiplied by its initial velocity (v1). The initial momentum of m2 is zero as it is stationary. After the collision, the final momentum of m1 can be calculated by multiplying its mass (m1) by its final velocity (v1'). Similarly, the final momentum of m2 can be calculated by multiplying its mass (m2) by its final velocity (v2'). The equation for conservation of momentum can be written as m1v1 + m2v2 = m1v1' + m2v2'. Since m2 > m1, it can be assumed that m2 will have a lower velocity after the collision compared to m1. This is because the total momentum of the system must remain constant. Therefore, the velocity of m2 after the collision will be less than the velocity of m1 before the collision.

b) The approximate momentum of m1 after the collision can be calculated using the equation for conservation of momentum mentioned above. Rearranging the equation, we get m1v1' = m1v1 + m2v2 - m2v2'. Since m2 > m1, m2v2 - m2v2' will be a positive value, which means that m1v1' will be greater than m1v1. This indicates that the final velocity of m1 (v1') will be greater than its initial velocity (v1). Therefore, the approximate momentum of m1 after the collision will be greater than its initial momentum. This can also be justified mathematically by considering the fact that kinetic energy is also conserved in an elastic collision. The equation for conservation of kinetic energy can be written as 1/2m1v1^2 = 1/2m1v1'^2. Since m1 is constant, this means that v1'^2 > v1^2, which further supports the idea that the final momentum of m1 will be greater than its initial momentum.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top