Calculating Impulse and Velocity in Inelastic Collision

AI Thread Summary
In the inelastic collision between block A (2.1 kg) moving at 7.8 m/s and block B (3.4 kg) moving at 4.2 m/s, the final velocity of the combined blocks after the collision is calculated to be 0.382 m/s to the right. The impulse experienced by block B on block A during the collision can be determined using the change in momentum formula, J = Δp. The discussion highlights confusion regarding which initial velocity to use, emphasizing the importance of understanding initial and final velocities in inelastic collisions. Participants suggest starting with known values and applying the impulse-momentum principle to find the solution. The conversation underscores the complexities of analyzing collisions involving objects with different speeds.
DoctorB2B
Messages
15
Reaction score
0

Homework Statement


Block A (2.1kg) is moving at 7.8 m/s to the right and collides head on with block B (3.4kg), which is moving at 4.2 m/s to the left.

a) What is the velocity (magnitude and direction) of the blocks after the collision, if the two blocks stick together?

b) What is the impulse (magnitude and direction) of block B on block A during the collision?


Homework Equations


m1v1 + m2v2 = (m1 +m2)(vf)



The Attempt at a Solution


a)I plugged in known values ((2.1)(7.8)+(3.4)(-4.2))=(5.5)vf
vf=0.382 to the right, since the number was positive

b)I'm not sure where to start.
 
Physics news on Phys.org
DoctorB2B said:

Homework Statement


Block A (2.1kg) is moving at 7.8 m/s to the right and collides head on with block B (3.4kg), which is moving at 4.2 m/s to the left.

a) What is the velocity (magnitude and direction) of the blocks after the collision, if the two blocks stick together?

b) What is the impulse (magnitude and direction) of block B on block A during the collision?

Homework Equations


m1v1 + m2v2 = (m1 +m2)(vf)

The Attempt at a Solution


a)I plugged in known values ((2.1)(7.8)+(3.4)(-4.2))=(5.5)vf
vf=0.382 to the right, since the number was positive

b)I'm not sure where to start.

Maybe start with what you know impulse is?
 
J=delta p

I wasn't sure where to start because I'm dealing with two objects moving at different speeds. I think I'm suppose to be dealing with initial and final velocities, however I don't know which initial velocity I would use since this is an inelastic collision.
 
DoctorB2B said:
J=delta p

I wasn't sure where to start because I'm dealing with two objects moving at different speeds. I think I'm suppose to be dealing with initial and final velocities, however I don't know which initial velocity I would use since this is an inelastic collision.

But they ask you what the impulse is on Block A. You know its initial p. And since matter wasn't destroyed that same mass of A now has acquired a final velocity, and hence has experienced a Δ in p right?
 
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 '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