A block that has a mass M = 4.25 kg is hanging at rest on a light string.A projectile with a mass m = 250 g moving horizontally strikes M and sticks to it.The block and its projectile swing up and the center of mass rises a distance h = 12.0 cm.The speed of the projectile is approximately
A)28 m/s
B)42 m/s
C)0.28 km/s
D)26 m/s
E)14 m/s

Question

Quizplus · Accepted Answer

The conservation of momentum and energy principles are used to solve this problem. Initially, the projectile has kinetic energy, which is converted into potential energy when the block and projectile system swings upwards. The conservation of momentum gives us the initial velocity of the combined mass right after the collision, and the conservation of energy allows us to calculate the height to which the system rises.Using the conservation of energy:$$ \frac{1}{2}(M + m)v^2 = (M + m)gh $$where $v$ is the velocity of the combined mass right after the collision, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the height (0.12 m). However, to find the initial velocity of the projectile, we first need to find $v$ using the conservation of momentum:$$ mv_{projectile} = (M + m)v $$Given:- $M = 4.25$ kg- $m = 0.25$ kg (since 250 g = 0.25 kg)- $h = 0.12$ mRearranging the momentum equation to solve for $v_{projectile}$, and knowing that the final velocity $v$ can be found through the energy conservation equation, we can solve for $v_{projectile}$.The calculations involve substituting the given values into the equations and solving for $v_{projectile}$, which yields approximately 28 m/s. This process involves using the given mass and height to first find the velocity of the combined system right after the collision using energy principles, and then using the momentum conservation principle to find the initial speed of the projectile.

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