On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with...

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

Solutions

Expert Solution

(a) What is the total kinetic energy of the two blocks after the collision (in joules)?

KE = (1/2)*0.2*12^2 + (1/2)*0.4*12^2 = 43.2 Jules

(b) What is the x-component of the total momentum of the two blocks after the collision?

x-component of the total momentum = 0.2*12*cos(30) + 0.4*12*cos(53.1) = 4.9605 kg m/s

(c)What is the y-component of the total momentum of the two blocks after the collision?

y-component of the total momentum = 0.2*12*sin(30) - 0.4*12*sin(53.1) = -2.6385 kg m/s

(a) What are the magnitude and direction of the velocity of the red glider after the collision?

0.2*8 = 0.2*(-2.2) + 0.6*velocity of the red glider after the collision

velocity of the red glider after the collision = + 3.4 m/s (towards right)

(b) yes collision is elastic

how far does the block travel horizontally while it is in the air?

vertical motion

s = ut + (1/2)*g*t^2

0.8 = 0 + 0.5*9.8*t^2

t = 0.4041 s

horizontal distance travelled = v * t = 4 * 0.4041 = 1.6162 m

genius_generous answered 2 weeks ago

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