1) On a horizontal frictionless surface, a small block with mass 0.200 kg has a
ID: 1772700 • Letter: 1
Question
1) 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 11.2 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.)
2) On a frictionless air track, a blue glider with mass 0.200 kg is moving to the right with a speed of 8.00 m/s. It strikes a red glider that has mass 0.600 kg and that is initially at rest. After the collision, the blue glider is moving to the left with a speed of 3.00 m/s.
(a) What are the magnitude and direction of the velocity of the red glider after the collision?
(b) Is this collision elastic?
3) A small block has a horizontal velocity of 4.00 m/s as it slides off the edge of a table. The table is a vertical distance of 1.15 m above the floor. If g = 9.80 m/s2, how far does the block travel horizontally while it is in the air?
Explanation / Answer
given,
on a horizontal frictionless track
mass m1 = 0.2 kg collides with m2 = 0.4 kg
After collision,
block 1 moves with v1 = 12 m/s in 30 deg north of east
block 2 moves with v2 = 11.2 m/s in 53.1 deg south of east
let i and j be unit vectors in east and north directions respectively
a. after collision total KE of two blocks
KE = 0.5m1v1^2 + 0.5m2v2^2
KE = 0.5*0.2*12^2 + 0.5*0.4*11.2^2
KE = 39.488 J
b. x component of the total momentum of the two blocks after collision
px = m1v1cos(30) + m2v2cos(53.1)
px = 0.2*12*cos(30) + 0.4*11.2*cos(53.1)
Px = 4.7683 kg m/s
c. y component of total momentum, Py = m1v1sin(30) - m2v2sin(53.1)
Py = 0.2*12*sin(30) - 0.4*11.2*sin(53.1)
Py = -2.38 kg m/s
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