NAME: 06: Two crates are moving of 16kg. Crate 1 is initially moving East at 4m/

Question

NAME: 06: Two crates are moving of 16kg. Crate 1 is initially moving East at 4m/sec ing towards each other. Crate 1 has a mass of 8kg and Crate 2 has a mass on, Crate 2 is observed to have stopped. The collision lasts for 0.1sec. Fill and Crate 2 is moving at 4m/sec West. They MEASUREMENT INSIDE THE MATRIX AS THEY ARE LISTED and round to NO MORE THAN 1 decimal. DO NOT PUT UNITS OF AT THE TOP.-30 points Before After impulse N-s (m/s/s) F object m O.24 -32 total 2A KO Power P m v P:()04)(16) 4) 2 otal

Explanation / Answer

TABLE 1:

Column 3 (v'):

v'2 = velocity of crate 2 = 0 (given in question)

now INitial momentum = Final momentum

m1v1 + m2v2 = m1v'1 + m2v'2

8*4 - 16*4 = 8*v + 0

v1' = -4m/s

Column 4: Already filled

Column 5:

p'1 = 8*-4 = -32

p'2 = 0

Total = p'1 + p'2 = -32

Column 6:

v1 = vfinal - vinitial = -4 - 4 = -8 m/s

v2 = 0- (-4) = 4 m/s

Column 7:

p1 = m(v1) = 8*-8 = -64

p2 = 16*4 = 64

Column 8-11:

Force = p/t

F1 = p1/0.1 = -64/0.1 = -640 N

F2 = p2/0.1 = 640 N

F = F1+F2 = 0

a1 = F1/m1 = -640/8 = -80 m/s^2

a2 = F2/m2 = 640/16 = 40 m/s^2

a = a1 + a2 = -40 m/s^2

Impulse I = Ft

I1 = -640*0.1 = -64 Ns

I2 = 640*0.1 = 64 Ns

I1+I2 = 0

TABLE 2:

KE1(j) = 0.5*mv^2 = 0.5*8*4*4 = 64 J

KE2(j) = 0.5*16*4*4 = 128 J

KEtotal = 128 + 64 = 192 J

KE'1(j) = 0.5*8*(-4)*(-4) = 64 J

KE'2(j) = 0.5*8*(0)*(0) = 0 J

KE1 = 0

KE2 = 128 J

KE = 128 J

Work done by crate 1 W1 = KE1 = 0

Work done by crate 2 W2 = KE2 = 128 J

Total Work done = KE = 128J

Power = Work done / time

P1 = 0

P2 = 128/0.1 = 1280 W

P = P1+P2 = 1280 W

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