A block is moving on a horizontal surface at a speed of 5.35 m/s strikes an iden

Question

A block is moving on a horizontal surface at a speed of 5.35 m/s strikes an identical stationary blocka glancing blow. After the collision, one block is found to be moving at a speed of 1.20 m/s in a direction making a 64.5 degree angle with the original line of motion. Assume that the blocks slide on the making a 64.5 degree angle with the original line of motion. Assume that the blocks slide on the horizontal surface without friction. (a) What is the velocity of the other block after the collision? Give magnitude and direction. (b) Given these data, determine whether the collision is elastic or inelastic: Calculate how much of the initial kinetic energy was lost (if any) in the collision.

Explanation / Answer

Using coonservation of momentum

momentum before collision =momentum after collision

let m be the mass of each block

then

m*5.35 = m*1.2*cos(64.5) + m*vx

vx = 4.83 m/s

and also

0 = m*1.2*sin(64.5) + m*vy

vy = -1.083 m/s

magnitude is v = sqrt(vx^2+vy^2) = sqrt(4.83^2+1.083^2) = 4.95 m/s

direction is theta = tan^-1(-1.083/4.83) = -12.63 degrees

below the 12.6 degrees to the original direction

B) initial kinetic energy is Ki = 0.5*m*u^2 = 0.5*m*5.35^2 = 14.3*m J

final KE = 0.5*m*1.2^2 + 0.5*m*4.95^2 = 12.97 *m

change in KE = 12.97*m - 14.3*m = -1.33*m

So the collision is inelastic collision

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