A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. A

Question

A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.80 m/s, at an angle of 25.0 with respect to the original line of motion. (Enter the direction with respect to the original line of motion. Include the sign of your answer. Consider the sign of the first ball's angle.)

a) Find the velocity (magnitude and direction) of the second ball after collision, in m/s and degrees, respectively.

b) Was the collision inelastic or elastic?

Explanation / Answer

if the velocity of the second ball after collision is V2 at an angle under the positive x-axes:
m*(5.3 m/s)=m*(4.8 m/s)*cos(25°)+m*V2*cos() (momentum coservation along the x-axes).

V2*cos()=0.9497................(1)

0=m*(4.8 m/s)*sin(25°)-m*V2*sin() (... coservation along the y-axes).
V2*sin()=2.028568 ................(2)

solving (1) & (2) gives:

V2=2.23987 m/s
=64.9°

if the kinetic energy conserved the collision is elastic, if not its inelastic.since the balls have the same mass we can only check if:

5.3^2=(4.8)^2+(2.23987)^2
its elastic collision.

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