How would you do C? Please explain the steps too not just put the answer. 3. In

Question

How would you do C? Please explain the steps too not just put the answer.

3. In the diagram above, blocks m and M are initially at rest. Block m slides down a smooth inclined plane and collides with block M. The blocks stick together and slide a distance x on a rough surface (with a coefficient of friction k) where they come to a stop. Express you answers in term of m, M, o, d, g and p () (2 points) Find the velocity of block m just before it collides with block M (b) (2 points) Find the velocity of the combined mass right after the collision. (c) (3 points) Find the amount of kinetic energy lost in the collision. (d) (3 points) Find the distance x.

Explanation / Answer

a) velocity block m just before it coolides with block M,

u = sqrt(2*g*d*sin(theta))

b) let v is the combined velocity after the collsion.

use conservation of momentum

m*u = (m + M)*v

==> v = m*u/( m + M)

= m*sqrt(2*g*d*sin(theta))/(m + M)

c) loss of kinetic energy in the collision = (1/2)*m*u^2 - (1/2)*(m + M)*v^2

= (1/2)*m*(2*g*d*sin(theta)) - (1/2)*(m + M)*(m*sqrt(2*g*d*sin(theta))/(m + M))^2

= (1/2)*m*(2*g*d*sin(theta)) - (1/2)*m^2*2*g*d*sin(theta)/(m + M)

= m*g*d*sin(theta)*(1 - m/(m + M))

= m*g*d*sin(theta)*(M/(m + M))

d) let x is the distance travelled before coming to stop.

Workdone by friction = change in kinetic energy

fk*x*cos(180) = 0 - (1/2)*(m + M)*v^2

-fk*x = -(1/2)*(m + M)*v^2

mue_k*(m + M)*g*x = (1/2)*(m + M)*(m*sqrt(2*g*d*sin(theta))/(m + M))^2

mue_k*g*x = m^2*g*d*sin(theta)/( m + M)^2

x = m^2*d*sin(theta)/(mue_k*( m + M)^2)

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