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
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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.
 
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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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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