A puck of mass m kg is placed inside a horizontal spring-loaded launcher. The sp

Question

A puck of mass m kg is placed inside a horizontal spring-loaded launcher. The spring constant is k N/m and the spring is compressed a distance d m. The spring is released and the puck leaves the launcher and slides forwards across the frictionless surface until it strikes a block of mass M kg at rest on the surface. In the following, give all answers in terms only of the given quantities m,k,d, and M.

a.) What was the puck's speed v, just before striking the block?

b.) After striking the block, the puck sticks to the block and they move together. What is the speed v12 of the puck-block combination after the collision?

c.) How much total kinetic energy (delta K) is lost in the collision?

d.) The system now explodes separating the puck and block. The puck emerges at rest. What is the speed of the block vB after this explosion?

Explanation / Answer

a) Apply conservation of energy

0.5*m*v^2 = 0.5*k*d^2

v = d*sqrt(k/m)

b) Apply conservation of momentum

m*v = (m + M)*v12

==> v12 = m*v/(m+M)

= m*d*sqrt(k/m)/(m + M)

= d*sqrt(k*m)/(m+M)

c) KEi = 0.5*m*v^2

= 0.5*k*d^2

KEf = 0.5*(m+M)*v12^2

= 0.5*(m+M)*(d*sqrt(k*m)/(m+M) )^2

= 0.5*k*d^2*(m/(m+M)

KEi - KEf = 0.5*k*d^2 - 0.5*k*d^2*(m/(m+M)

= 0.5*k*d^2*(1 - m/(m+M))

= 0.5*k*d^2*(M/(m+M))

d)

Again Apply conservation of momentum

M*vB = (m+M)*V12

vB = (m+M)*V12/M

= (m+M)*(d*sqrt(k*m)/(m+M))/M

= (d/M)*sqrt(k*m)

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