Physics 11: Calculate Velocity of Student Throwing Backpack on Ice Surface

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    Physics Physics 11
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A 55 kg student on a frictionless ice surface throws a 10 kg backpack at 8.3 m/s south. To find the student's velocity after the throw, the impulse-momentum principle is applied, stating that the impulse on the backpack equals the impulse on the student. The calculation involves using the equation m_student * v_student = m_backpack * v_backpack. The resulting velocity of the student is 1.5 m/s north. This approach confirms the correct application of physics principles in solving the problem.
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Homework Statement



A 55 kg student wearing skates stands at rest on a frictionless horizontal ice surface. She throws a 10 kg backpack horizontally with a speed8.3 m/s .
What is the velocity of the student after she throws the backpack?

Homework Equations



F\Deltat = m\Deltav

The Attempt at a Solution


so the answer should be 1.5 m/s [N] and i got this by dividing 10 kg (backpack) by 55 kg (student) and then multiplying that answer by 8.3 m/s . is this the right way or is there a proper formula i should follow?
thanks !
 
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That's correct. The impulse given to the backpack will be exactly equal and opposite to the impulse given to the student. Thus:
mstudent*vstudent = mbackpack*vbackpack

Just solve for vstudent.
 
great, thanks ! :smile:
 
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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