A billiard player sends the cue ball toward a group of three balls that are init

Question

A billiard player sends the cue ball toward a group of three balls that are initially at rest and in contact with one another. After the cue ball strikes the group, the four balls scatter, each traveling in a different direction with different speeds as shown in the figure below. If each ball has the same mass, 0.16 kg, determine the total momentum of the system consisting of the four balls immediately after the collision. (Assume v1 = 0.33 m/s, 1 = 70°, v2 = 0.55 m/s, 2 = 30°, v3 = 0.21 m/s, v4 = 0.48 m/s.)

magnitude:

direction: ° counterclockwise from the +x axis

Explanation / Answer

let the unit vector along x axis be i and unit vector along y axis be j.

then in vector notation, v1=0.33*(cos(theta1) i + sin (theta1) j)=0.11287 i + 0.31 j m/s

v2=0.55*(cos(theta2) i - sin(theta2) j)=0.4763 i -0.275 j m/s

v3=-0.21 j m/s

v4=-0.48 i m/s

as momentum=mass * velocity

total momentum of the system=m1*v1+m2*v2+m3*v3+m4*v4

but as given that , m1=m2=m3=m4=0.16 kg

total momentum=0.16*(v1+v2+v3+v4)

=0.16*(0.11287 i + 0.31 j + 0.4763 i -0.275 j -0.21 j -0.48 i)=0.017467 i -0.028 j kg.m/s

magnitude=sqrt(0.017467^2+0.028^2)=0.033 kg.m/s

angle with x axis=-58.043 degrees

that is angle with x axis is 58.043 degrees in clockwise direction or -58.043 degrees in anticlockwise direction.

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