A ball with mass M, moving horizontally at 4.00 m>s, collides elastically with a block with mass 3M that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit.

- When two objects collide with no external force acting on the system the linear momentum of the system is conserved. The initial (Pi) and final (Pf) linear momentum are equal:

Pi = Pf

M*vi = M*vf + 3M*Vf

vi = vf + 3*Vf

4 = vf + 3*Vf

- For elastic collision between two particles the relative velocities before and after collision have the same magnitude but opposite sign; so,

vi - 0 = Vf - vf

4 = Vf - vf

- Solve the above two equation simultaneously.

8 = 4*Vf

Vf = 2 m/s

vf = -2 m/s

- When the ball hits the object it swing under the influence of gravity only. Hence, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.

ΔK.E = ΔP.E

0.5*(3M)*Vf^2 = (3M)*(g)*(h)

h = Vf^2 / 2*g

- Plug in the values:

h = 2^2 / 2*9.81

h = 0.2039 m

- We can see that the maximum angle can be given as θ according trigonometric relation as follows:

θ = arccos [ ( L - h ) / L ]

θ = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]

θ = 53.7°

tallinn tallinn · Answer 2 · 2022-03-23T12:50:54+01:00

The maximum angle through which the block swings after it is hit θ is = 53.7°

Calculation of Mass

Given:

The mass of ball is = M

The mass of object is = 3M

The wire length L is = 0.5 m

The velocity of ball vi is = 4.0 m/s

Then The velocity of ball vf

After that The velocity of object Vf

Now we Find:

Find the maximum angle through which the block swings after it is hit that is :

When two objects collide with no external force acting on the system the linear momentum of the system is conserved. Then The initial (Pi) and also final (Pf) linear momentum are equal:

Pi is = Pf

M*vi is = M*vf + 3M*Vf

vi is = vf + 3*Vf

4 is = vf + 3*Vf

Now For elastic collision between two particles the relative velocities before and also after collision have the same magnitude but opposite signs; so,

vi - 0 is = Vf - vf

4 is = Vf - vf

Solve the above two equations simultaneously.

8 is = 4*Vf

Vf is = 2 m/s

vf is = -2 m/s

When the ball hits the object it swings under the influence of gravity only. Hence proof, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.

ΔK.E is = ΔP.E

0.5*(3M)*Vf^2 is = (3M)*(g)*(h)

h is = Vf^2 / 2*g

Then we Plug in the values is:

h is = 2^2 / 2*9.81

h is = 0.2039 m

Now We can see that the maximum angle can be given as θ according trigonometric relation as follows:

θ is = arccos [ ( L - h ) / L ]

θ is = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]

θ is = 53.7°

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