A billiard ball is moving in the x-direction at 4.0 m/s andstrikes another billi

Question

A billiard ball is moving in the x-direction at 4.0 m/s andstrikes another billiard ball moving in the y direction at 1.5 m/s.After the collision, the second ball stops. Two balls haveidentical mass. What is the % change in kinetic energy of the system as aresult of the collision? A billiard ball is moving in the x-direction at 4.0 m/s andstrikes another billiard ball moving in the y direction at 1.5 m/s.After the collision, the second ball stops. Two balls haveidentical mass. What is the % change in kinetic energy of the system as aresult of the collision?

Explanation / Answer

      m1 *u1   +   m2*u2   =   m1*v1  +   m2* v2    Since second ballstops,   v2   =   0and mass is same    =>   m *u1   +   m*u2   =   m*v1         or      v1   =   u1   +   u2    ( Bold face letters are vectors)    vector addition is iven by (magnitude)       v12   =   u12   +   u22   +   2* u1 * u2 * cos       =   anglebetween direction of u1 and u2   =   900       v12   =   4.02   +   1.52   +   2* 4.0 * 1.5 * cos900   =   18.25      m2    Total initialenergy   Ki=   (1/2) * m *u12   +   (1/2)* m *u22   =   0.5 *m * (4.02   +   1.52)   =   18.25* 0.5 * m    Total finalenergy   Kf   =   (1/2)* m *v12   =   0.5 *m * 18.25    Change in energy%   =  {( Kf -Ki )/ Ki} *100%   =   0 %    Total initialenergy   Ki=   (1/2) * m *u12   +   (1/2)* m *u22   =   0.5 *m * (4.02   +   1.52)   =   18.25* 0.5 * m    Total finalenergy   Kf   =   (1/2)* m *v12   =   0.5 *m * 18.25    Change in energy%   =  {( Kf -Ki )/ Ki} *100%   =   0 %    

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