A block of wood of mass m1 = 1.00 kg slides to the right on a frictionless horiz

Question

A block of wood of mass m1 = 1.00 kg slides to the right on a frictionless horizontal table with a speed of 30.00 m/s. It collides with a second piece of wood of mass m2 = 2.00 kg which is stationary at position X. The collision is completely inelastic. After the collision the wood slides across a rough horizontal surface ( ? = 0.30) from B to C for a distance of 10.00 m. After position C, the surface becomes smooth and the wood collides with a spring with spring constant of k = 30.00 N/m.

(a)How much is the spring compressed by the wood?

Consider now that the spring pushes the wood back in the opposite direction.

(b) At what locations along the horizontal plane is mass m2 stationary?

Explanation / Answer

a) after the complete inelastic collision, mass of the combined object will be (1+2)=3kg
so from momentum conservation, resulted velocity of the combined object is = 1*30/3 = 10m/s

from B to C, the friction coefficient of the surface is ( = ) 0.3

so accelaration = -mg/m = -g = -0.3*10 = -3m/s2

so at C velocity of the object will be v = (u2 + 2as) = (102 - 2*3*10) = 40 m/s (s=10, distance from B to C)

now, if the spring compressed x mt. for the collision, them from energy conservation,

mv2/2 = kx2/2

or, mv2 = kx2

or, 3*40 = 30*x2

or, 4 = x2

or, x = 2

So, the spring will be compressed by 2 mt. by the wood

b) now the object will regain its velocity and rebound. So at C it velocity will be same as v(= 40)

if it stops after s distance towards B, then from v2 = u2 - 2as

s = (u2/2a) = (40/2*3) = 2.58 mt. from C (a = -3m/s2 calculated above, and u = 40 m/s)

Cheers...

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