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Conceptual Physics 
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#1
Mar1109, 02:29 PM

P: 5

Anybody in here taking Physics class right now or in the past? Can you help me out...i have 4 more questions:
1) How much impulse stops a 50kg carton sliding at 4 m/s when it meets a rough surface? Conservation of momentum: mv(befor) = mv(after) 2) If a Mack truck and Ford Escort have a head on collision, which vehicle will experience the greater force of impact? the greater impluse? the greater change in momentum? the greater deceleration? 3)When a stationary uranium nucleus undergoes fission, it breaks into two unequal chunks that fly apart. What can you conclude about the momenta of the chunks? what can you conclude about the speed of the chunks? 4) Judy (mass 40.0 kg) standing on slippery ice, catches her leaping dog, Atti (mass 15kg), moving horizontally at 3.0 m/s. What is the speed of judy and her dog after the catch? 


#2
Mar1109, 02:46 PM

Mentor
P: 40,707

You need to show some work before we can offer tutorial help. Show us the relevant equations (the stuff you are learning), and show us how you can apply them to these questions.... 


#3
Mar1109, 02:46 PM

P: 5

Here is what i did but i dont know if it's correct....
1) Impulse = mass * delta(velocity) = m*dv = d(mv) m = 50 kg dv= 4  0 = 4 I = 50,000 * 4 = 200,000 2)  F=ma => and they both experience the same  Once again, I = d(mv), assuming they both have the same difference in velocity before and after, mack truck experiences greater impulse  Impulse is the change in linear momentum... same as the last one.  Ford experiences greater deceleration. F1 = F2 => m(mack)*a(mack) = m(ford)*a(ford). Since m(ford) << m(mack), a(ford) must be >> a(mack). 3) Conservation of momentum & perfectly inelastic collision (explosion) => (M+m)*v(0) = M*v(1) + m*v(2) The sum of the two piece's momenta is equal to the initial momentum of the uranium nucleus. The chunk with a higher mass (M) will have a smaller speed than the chunk with lower mass (m). 4) Same as number three, this time the equation is reversed M(judy) * v(judy) + m(dog)*v(dog) = (M+m)*v(final) (40)*0 + (15)*(3) = (40+15)*v(f) => v(f) = 0.818 m/s 


#4
Mar1109, 02:49 PM

Mentor
P: 40,707

Conceptual Physics
Much better! Most looks right, except on #2. Think about what the velocities are going to be like after the collision, and hence the deltaVs, which go towards the accelerations....



#5
Mar1109, 03:12 PM

P: 5

thanks for your help



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