A 0.500 kg sphere moving with a velocity given by strikes another sphere of mass

Question

A 0.500 kg sphere moving with a velocity given by

strikes another sphere of mass 1.50 kg moving with an initial velocity of

(a) The velocity of the 0.500 kg sphere after the collision is

()ˆi+()ˆj+()ˆk

m/s

Identify the kind of collision (elastic, inelastic, or perfectly inelastic).

Choose:

elastic

inelastic    

perfectly inelastic

(b) Now assume the velocity of the 0.500 kg sphere after the collision is

m/s

Identify the kind of collision.

elasticinelastic    perfectly inelastic

(c) Take the velocity of the 0.500 kg sphere after the collision as

Find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a positive value and a negative value. Report each with their corresponding final velocities.)

a (positive value)      v2f = m/s a (negative value)      v2f = m/s

Explanation / Answer

Using law of conservation of linear momentum

momentum before collision = momentum after collision

0.5*(2i-2.5j+1k) + 1.5*(-1i+2j-3.4k) = 0.5*(-0.6i+3j-8k) + 1.5*(vxi +vy j + vz k)

-0.5i + 1.75 j - 4.6 k = -0.3 i + 1.5 j - 4k + (1.5Vx i )+(1.5vy j) + (1.5*vz k)

comparing on both sides

-0.5 = -0.3+(1.5*vx)

vx = -0.13 m/sec

1.75 = (1.5+(1.5*vy)

vy = 0.167 m/sec

-4.6 = -4+(1.5*vz)

vz = -0.4 m/sec

so final velocity of 1.5 kg is

V = -0.13 i + 0.167 j - 0.4 k

initial kinetic energy is KEi = 0.5[0.5*(2^2+2.5^2+1^2) + 1.5*(1^2+2^2+3.4^2)] = 15.2325 J

Final kinetic energy is KEf = 0.5*0.5*(0.6^2+3^2+8^2) +(0.5*1.5*(0.13^2+0.167^2+0.4^2) = 18.5 J

since KEi is not equal to KEf then the collision is inelastic collision

b) Using law of conservation of linear momentum

momentum before collision = momentum after collision

0.5*(2i-2.5j+1k) + 1.5*(-1i+2j-3.4k) = 0.5*(-0.25i+0.875j-2.3k) + 1.5*(vxi +vy j + vz k)

-0.5i + 1.75 j - 4.6 k = (-0.125 i + 0.4375 j - 1.15k) + (1.5Vx i )+(1.5vy j) + (1.5*vz k)

comparing on both sides

-0.5 = -0.125+(1.5*vx)

vx = -0.25 m/sec

1.75 = (0.4375+(1.5*vy)

vy = 0.875 m/sec

-4.6 = -1.1+(1.5*vz)

vz = -2.34 m/sec

so final velocity of 1.5 kg is

V = -0.25 i +0.875 j - 2.34 k

initial kinetic energy is KEi = 0.5[0.5*(2^2+2.5^2+1^2) + 1.5*(1^2+2^2+3.4^2)] = 15.2325 J

Final kinetic energy is KEf = 0.5*0.5*(0.25^2+0.875^2+2.3^2) +(0.5*1.5*(0.25^2+0.875^2+2.34^2) = 6.25 J

since KEi is not equal to KEf then the collision is inelastic collision

C)

Using law of conservation of linear momentum

momentum before collision = momentum after collision

0.5*(2i-2.5j+1k) + 1.5*(-1i+2j-3.4k) = 0.5*(-1i+3.5j+ak) + 1.5*(vxi +vy j + vz k)

-0.5i + 1.75 j - 4.6 k = (-0.5 i + 1.75 j +0.5ak) + (1.5Vx i )+(1.5vy j) + (1.5*vz k)

comparing on both sides

-0.5 = -0.5+(1.5*vx)

vx = 0 m/sec

1.75 = (1.75+(1.5*vy)

vy = 0 m/sec

-4.6 = (0.5a)+(1.5*vz)

vz = (-4.6-(0.5a))/1.5

if the colliision is elastic then

initial kinetic energy is KEi = 0.5[0.5*(2^2+2.5^2+1^2) + 1.5*(1^2+2^2+3.4^2)] = 15.2325 J

Final kinetic energy is KEf = 0.5*0.5*(1^2+3.5^2+a^2) +(0.5*1.5*(0^2+0^2+((-4.6-(0.5a))/1.5)^2)

15.2325 J = 0.5*0.5*(1^2+3.5^2+a^2) +(0.5*1.5*(0^2+0^2+((-4.6-(0.5a))/1.5)^2)

a = -6.76

and a = 2.16

final velocity of 1.5 kg is 0 i + 0j + (-4.6-(0.5*2.16))/1.5 k = -3.79 k

or (-4.6+(0.5*6.76))/1.5

vz = -0.82 m/sec

so another value of final velocity of 1.5 kg is -0.82 k

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