Two blocks are sliding on a horizontal frictionless surface with velocities show

Question

Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.550 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.

block A is moving to the right at .2 m/s

block B is moving upwards at .4 m/s

a) What is the speed of the combined blocks after the collision?
(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?
(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Explanation / Answer

here,

mass of block A = ma = 0.5 kg
mass of block B = mb = 0.55 kg

Va = 0.2 m/s
Vb = 0.4 m/s

Part A:
For perfctly in elastic collisoin
momentum beore collision = momentum beore collision
ma*va +mb*vb = (ma+mb)V
(0.5*0.2)i + (0.55*0.4)j = (0.5+0.55) *V
V = (0.1i + 0.22j) 1/ 1.05
V = (0.09i + 0.20j)

Vnet = sqrt (x^2 + y^2)
Vnet = sqrt (0.09^2 + 0.20^2)
Vnet = 0.219 m/s

the speed of the combined blocks after the collision is 0.219 m/s

Part B:

tanA = Vy/Vx = 0.20/0.09
A = arcTan(2.22)
A = 65.75 Degrees

angle made by velocity of the combined blocks with the +x-axis after the collision is 65.75 degrees

Part B:
KEi = 0.5 *ma*va^2 + 0.5*mb*vb^2
KEi = 0.5 * 0.5 * 0.2 + 0.5 * 0.55 * 0.4^2
KEi = 0.094 J

KEf = 0.5 *(ma+mb)*v^2
KEf = 0.5 * (0.5+0.55) * 0.219^2
KEf = 0.025 J

Decreased Energy = 0.094 - 0.025
Decreased Energy = 0.069 J

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