Work and Kinetic Energy and elastic collision

AI Thread Summary
In an elastic collision involving a 2 kg mass that continues moving at one-fourth its original speed, the key to finding the mass of the struck body lies in applying the conservation of momentum and kinetic energy principles. The relationship between the velocities before and after the collision indicates that the speed of approach equals the speed of separation. By setting up the equation based on the ratio of velocities and rearranging terms, it becomes clear that the mass of the second body can be derived from the equation 0 = 3(2 kg) - 5m2. Solving this yields the mass of the struck body, providing a clear path to the solution. The discussion emphasizes the importance of understanding the principles of elastic collisions in solving such problems.
siemieniuk
Messages
7
Reaction score
0
hey all... have a question:

a body of 2 kg mass makes an elastic collision with another body at rest and afterwards continues to move in the original direction but with one-fourth of it's original speed. What is the mass of the struck body?

I started with comparing kinetic energy, then took a look at momentum... but I'm just not sure how I can solve for body #2's mass when I don't have a resultant velocity or anything... please point me in the right direction.

Thanks
 
Physics news on Phys.org
But you do have a resultant velocity, as it is elastic. Therefore, speed of approach must equal the relative speed after the collision.

V = Z - W

Where V is the initial velocity, Z is the resulting velocity of the struck object, and W is the resulting velocity of the first object.
 
Let V be final velocity and v be initial velocity of the 2 kg mass. Take the basic equation and multiply each side by v:

V/v = [(m1-m2)/(m1 + m2)]

We know that the ratio of V/v = 1/4. Therefore

1/4 = (m1 - m2)/(m1 + m2)

Cross multiply:

m1 + m2 = 4m1 - 4m2

0 = 3m1 - 5m2

0 = 3 (2 kg) - 5 m2
 
Sorry; I meant divide each side by v
 
thanks... appreciate it more than you know
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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 hanging 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...
Back
Top