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On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediately after the collision, the 0.200 kg block is moving at 12.0 m/s in the direction 30° north of east and the 0.400 kg block is moving at 11.6 m/s in the direction 53.1° south of east. Use coordinates where the +x-axis is east and the +y-axis is north.
(a) What is the total kinetic energy of the two blocks after the collision (in joules)?(b) What is the x-component of the total momentum of the two blocks after the collision? (Indicate the direction with the sign of your answer.)(c) What is the y-component of the total momentum of the two blocks after the collision? (Indicate the direction with the sign of your answer.)

Respuesta :

Answer:

Part a)

[tex]K = 41.31 J[/tex]

Part b)

[tex]P_x = 4.86 Ns[/tex]

Part b)

[tex]P_y = -2.51 Ns[/tex]

Explanation:

Part a)

Kinetic energy of two blocks is given as

[tex]K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2[/tex]

[tex]K = \frac{1}{2}(0.200)(12^2) + \frac{1}{2}(0.400)(11.6)^2[/tex]

so we will have

[tex]K = 41.31 J[/tex]

Part b)

X component of total momentum of two blocks is given as

[tex]P_x = m_1v_{1x} + m_2v_{2x}[/tex]

[tex]P_x = 0.200(12 cos30) + 0.400(11.6cos53.1)[/tex]

[tex]P_x = 2.08 + 2.78[/tex]

[tex]P_x = 4.86 Ns[/tex]

Part b)

Y component of total momentum of two blocks is given as

[tex]P_y = m_1v_{1y} + m_2v_{2y}[/tex]

[tex]P_y = 0.200(12 sin30) - 0.400(11.6sin53.1)[/tex]

[tex]P_y = 1.20 - 3.71[/tex]

[tex]P_y = -2.51 Ns[/tex]

The total momentum is "4.82 [tex]\frac{Kgm}{s}[/tex]" towards the east and towards the south "-2.45 [tex]\frac{Kgm}{s}[/tex]".

Momentum:

[tex]m_1 = 0.2\ Kg\\\\m_2 = 0.4\ Kg\\\\v_1 = 12.0\ \frac{m}{s} \ at \ 30^{\circ}[/tex]  north of east

[tex]v_2 = 11.4 \ \frac{m}{s} \ at \ 53.1^{\circ}[/tex] south of east

Calculating the total momentum of the X-component:

[tex]\to m_1 \times v_1 + m_2\times v_2\\\\ \to 0.2 \times 12.0 \times \cos(30^{\circ}) + 0.4 \times 11.4 \times \cos(53.1^{\circ})\ \frac{Kgm}{s}\\\\\to 4.82\ \frac{Kgm}{s}\ \ \text{}\\\\[/tex]Towards East

Calculating the total momentum of the Y-component:

[tex]\to m_1 \times v_1 + m_2 \times v_2\\\\\to 0.2\times 12.0\times \sin(30^{\circ}) - 0.4\times 11.4\times \sin(53.1^{\circ})\ \frac{Kgm}{s}\\\\\to -2.45 \ \frac{Kgm}{s}\ \text{}[/tex]Towards South

So, the final answer for the total momentum is "4.82 [tex]\frac{Kgm}{s}[/tex]towards the east and towards the south "-2.45 [tex]\frac{Kgm}{s}[/tex]".

Find out more information about the momentum here:

brainly.com/question/4956182