Physics, 04.12.2019 00:31 tiwaribianca475

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. initially ball a is moving upward along the y axis at 2.0 m/s and ball b is moving to the right along the x axis with speed 3.7 m/s. after the collision ball b is moving along the positive y axis. a. what is the speed of ball a and b after the collision? b. what is the direction of motion of ball a after the collision? c. what is the total momentum and kinetic energy of the two balls after the collision?

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Answer from: elishane864

The momentum conservation allows to find the results for the questions about the collision of the balls are:

A) The velocities after the collision are: vₐ = 3.7 i m/s and v_b = 2 j m/s

B) The direction of ball A is: x-axis

C) The total moment is: $p_{total}$ = 4.2m and the direction is θ = 28.4º

The total kinetic energy is: $K_{total}$ = 8.84m J

Given parameters

Ball A initially moves y-axis at $v_{oay}$ = 2.0 m / s Ball B moves end x axis v_[obx}

vobx = 3.7 m / s Ball B moves on the y-axis after collision

To find

A) The speeds after the collision.

B) direction of the ball A.

C) The moment and energy after the crash.

Momentum is defined by the product of mass and velocity. When a system is isolated, there is no external force, the momentum is conserved. This is an important conservation law of physics.

p = m v

Where the bold letters indicate vectors, m is the mass and v the velocity.

A) The system is formed by the two balls, therefore it is isolated and the momentum is conserved. Since the momentum is a vector quantity we write the components on each axis, see attached for a schematic.

Initial instant. Before the crash.

They indicate that ball B moves in the axis and after the collision.

x-axis

p₀ₓ = m $v_{obx}$ vobx

y-axis

$p_{oy} = m v_{oay}$

Final moment. After the crash.

x-axis

$p_{fx} = m v_{fax}$

y- axis

$p_{fy} = m v_{fay} + m v_{fby}$

The momentum is preserved

x- axis

p₀ₓ = $p_{fx}$

m $v_{obx}$ = m $v_{fax}$

$v_{fax} = v _{obx}\\v_{fax} = 3.7 m/s$

y- axis

$p_{oy} = p_[fy} \\m v_{oay} = m v_{fay} + m v_[fby}\\v_{oay} = v_{fay} + v_{fby}$

2 = $v_{fay} + v_fby}$

Ball A moves in the x-axis therefore it has no final velocity in the y-axis.

$v_{fay}=0$

$v_{fby}$ = 2 m / s

B) Ball A moves on the x-axis.

C) the total momentum is the sum of the momentum of each particle

$p_{total} = p_a+p_b$

To find the module let's use the Pythagorean theorem.

$p_{total} = m \sqrt{2^2 + 3.7^2}$

p = 4.2m

Let's use trigonometry for the direction.

tan θ = $\frac{p_{fb}}{p_{fa}}$

θ = tan⁻¹ $\frac{p_{fb}x}{p_{fa}}$

θ = tam⁻¹ $\frac{2}{3.7}$

θ = 28.4º

The total kinetic energy is

$K_{total}= K_1+K_2$

$K_{total}$ = ½ m v_a^2 + ½ m v_b^2

$K_{total}$ = $\frac{m}{2}$ (3.7² + 2²)

$K_{total}$ = 8.845 m

In conclusion using momentum conservation we can find the results for the questions about the collision of the balls are:

A) The velocities after the collision v_a = 3.7 i m/s and v_b = 2 j m/s

B) The direction of ball A is: x-axis

C) The total momentum is: $p_{total}$ = 4.2m and the direction is θ = 28.4º

The total kinetic energy is: K_ {total} = 8.84m J

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Answer from: 6clevare

Explanation:

Given

Initially A moves towards Y axis with a velocity of 2 m/s and B moves towards X axis with velocity of 3.7 m/s

mass of both the block is same let say m

Conserving momentum in x direction

$P_i_x=m\times 3.7$

P_f_x=m v_x

P_i_x= P_f_x

$m\times 3.7=m\times v$

v=3.7 m/s

thus A moves with a velocity of 3.7 m/s towards x axis in order to conserve the momentum in x direction

Thus ball B moves towards Y direction with v=2 m/s

And ball A moves towards X direction with v=3.7 m/s

(c)Total momentum $=\sqrt{P_x^2+P_y^2}$

$=\sqrt{(3.7m)^2+(2m)^2}$

=4.20 m

Kinetic Energy =K.E._a+K.E._b

$=\frac{1}{2}mv_a^2+\frac{1}{2}mv_b^2$

$=\frac{1}{2}m(3.7)^2+\frac{1}{2}m(2)^2$

=8.84 m J

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