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€ Chrome File Edit View History Bookmarks People Window Help 100% Sat 11:11 PM JJ5521 c Chegg C https:/ ne 654272 Q8 /www.webassign t/web/Student/Assignment-Responses/Mast?dept EE: Apps A&M; organizations TCC Miscellaneous PT WTC Lu Hot Schedules onli 11. 0/6 points I Previous Answers YF13 8. P.048 My Notes Ask Your Teacher A 10.0-g marble slides to the left with a velocity of magnitude 0.300 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 35.0-g marble sliding to the right with a velocity of magnitude o.150 m/s a) Find the velocity of each marble magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. Take right as the positive x direction.) 15 X m/s (smaller marble m/s (larger marble) b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. kg m/s (smaller marble) kg m/s (larger marble (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. (smaller marble) J (larger marble) 12. 2/2 points I Previous Answers YF13 8.P.061 My Notes Ask Your Teacher he M MM

Explanation / Answer

for elastic collisions

according to conservation of linear momentum

m1*u1 + m2*u2 = m1*v1+m2*v2

m1*(u1-v1) = m2*(v2-u2)........(1)

according to conservation of energy

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

m1*(u1^2 - v1^2) = m2*(v2^2-u2^2).....(2)

from 1 &2

u1 + v1 = u2+v2

u1 - u2 = v2 - v1

v2 = u1 - u2 + v1........(3)

3 in 1

m1*(u1-v1) = m2*(u1 - u2 + v1 - u2)

v1 = u1*(m1-m2)/(m1+m2) + 2*m2*u2/(m1+m2)

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

m1 = 10 kg     u1 = -0.3 m/s

m2 = 35 kg      u2 = 0.15 m/s

v1 = ?         V2 = ?

v1 = u1*(m1-m2)/(m1+m2) + 2*m2*u2/(m1+m2)

v1 = (-(0.3*(10-35))+(2*35*0.15))/(10+35)

v1 = 0.4 m/s ( smaller) <<<---answer

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

v2 = ((0.15*(35-10))-(2*10*0.3))/(10+35)

v2 = -0.05 m/s (larger)

part(b)

Psmaller = m*(v1-u1) = 0.01*(0.4-(-0.3)) = 0.007 kg m/s    <-----answer

Plarger = m2*(v2-u2) = 0.035*(-0.05+0.15) = 0.0035 kg m /s   <-----answer

part(c)

dKEsmaller = 0.5*m1*(v1^2-u1^2 )

dKE = 0.5*0.01*(0.4^2-0.3^2) = 0.00035 J   <-----answer

dKE larger = 0.5*m2*(v2^2-u2^2)

dKE = 0.5*0.035*(0.05^2-0.15^2) = 0.00035 J <-----answer

potential energy at height h = 100 feet = 30.48 m

PE = m*g*h

Kinetic energy of the kitten before hitting the ground = KE = 0.5*m*v^2

from energy conservation

KE = PE

0.5*m*v^2 = m*g*h

v = sqrt(2*g*h)

v = sqrt(2*9.8*30.48)

v = 24.44 m/s

v = 54.67 miles per hour <<<<--------------answer

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