A block having mass mB = 1 [kg] slides with speed vB = 2 [m/s] on a frictionless

Question

A block having mass mB = 1 [kg] slides with speed vB = 2 [m/s]
on a frictionless surface as shown below.
It collides with a second stationary block having mass mM = 3 [kg],
and the two blocks stick together.
The combined blocks then move together as a single unit
and compress an ideal spring of stiffness k = 4 [N/m] from it relaxed position.
(a) Use the conservation of momentum to determine the speed of the
combined blocks after the collision.
(b) Then use the conservation of energy to determine how far the spring is
compressed.

Explanation / Answer

The basic approach is to find out how fast the combined block and bullet are moving after the collision by using conservation of momentum. Then, use conservation of energy to calculate height by noting that the kinetic energy of the combined mass is transformed into gravitational potential energy. NOTE: Energy is not conserved due to the inelastic collision so you can't just convert the energy of the bullet into potential energy of block+bullet. mv + 0 = (m+M)v1, where v1 = velocity of block and bullet after collision v1 = mv / (m+M) v1 = .01 * 300 / (.01 + 1.0) v1 = 2.97 m/s 1/2 (m+M) * v1^2 = (m+M)gh h = v1^2 / 2g h = 2.97^2 / (2*9.8) h = 0.450 meters

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