In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at 1.10 m/s along a line making an angle of 29.0° with its original direction of motion, and the second ball has a speed ?

1 Answer
Jun 7, 2016

Velocity of 2nd ball after collision~~0.61m/s and
velocity of cue ball before collision ~~1.26m/s

Explanation:

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Let
Before collision he Cue ball is moving along the positive direction of X-axis and another ball is at rest lying at origin as shown in adjoining figure.

  • m->"the mass of each ball"
  • u_c->"Velocity of cue ball before collision"
  • u_a->"Initial velocity of another ball hit by cue ball"
  • v_c->"Velocity of cue ball after collision"
  • v_a->"velocity of another ball after collision"
  • alpha="Angle subtended by" v_c "with its initial direction of motion"
  • beta="Angle subtended by"v_a"with the initial direction of motion of cue ball.."

Given

v_c=1.1ms^-1 " ";u_a=0;" "alpha = 29^@

We are to find out the value of v_a

To solve this problem let us assume that the collision is perfectly elastic and both conservation of KE and Conservation of linear momentum are valid here.

Applying law conservation of KE we can write

1/2xxmxxu_c^2+1/2mxxu_a^2=1/2xxmxxv_c^2+1/2xxmxxv_a^2

=>u_c^2+u_a^2=v_c^2+v_a^2

Inserting u_a=0 we get " "u_c^2=v_c^2+v_a^2.....(1)

Now applying law of conservation of momentum

Along X-axis we can write

mxxu_c+ mxxu_a=mxxv_cxxcosalpha+mxxv_axxcosbeta

Inserting u_a=0 we have

u_c=v_cxxcosalpha+v_axxcosbeta.....(2)

Now applying law of conservation of momentum

along Y-axis we can write. ( there was no component of initial ovelocity along Y-axis)

0=mxxv_cxxsinalpha-mxxv_axxsinbeta

0=v_cxxsinalpha-v_axxsinbeta.....(3)

Now squaring eqation (2) and (3) and adding we get

u_c^2=v_c^2(cos^2alpha+sin^2alpha)+v_a^2(cos^2beta+sin^2beta)+2v_cv_a(cosalphacosbeta-sinalphasinbeta)

=>u_c^2=v_c^2+v_a^2+2v_cv_a(cosalphacosbeta-sinalphasinbeta)

Now combining equation(1) with this we get

2v_cv_a(cosalphacosbeta-sinalphasinbeta)=0

=>cos(alpha+beta)=0

=>alpha+beta=90

putting given value" "alpha=29^@ we get

29^@+beta=90=>beta=61^@

Now inserting the value of alpha,beta,v_c in equation(3)

we can write

1.1*sin29^@-v_a*sin61^@=0

:.v_a=(1.1*sin29^@)/(sin61^@)~~0.61m/s

The velocity of cue ball before collision

u_c=sqrt(v_c^2+v_a^2)=sqrt(1.1^2+0.61^2)~~1.26m/s