Question

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.)

Answer 1

Answer:

Part a)

$K = 41.31 J$

Part b)

$P_x = 4.86 Ns$

Part b)

$P_y = -2.51 Ns$

Explanation:

Part a)

Kinetic energy of two blocks is given as

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

$K = \frac{1}{2}(0.200)(12^2) + \frac{1}{2}(0.400)(11.6)^2$

so we will have

$K = 41.31 J$

Part b)

X component of total momentum of two blocks is given as

$P_x = m_1v_{1x} + m_2v_{2x}$

$P_x = 0.200(12 cos30) + 0.400(11.6cos53.1)$

$P_x = 2.08 + 2.78$

$P_x = 4.86 Ns$

Part b)

Y component of total momentum of two blocks is given as

$P_y = m_1v_{1y} + m_2v_{2y}$

$P_y = 0.200(12 sin30) - 0.400(11.6sin53.1)$

$P_y = 1.20 - 3.71$

$P_y = -2.51 Ns$