A sphere(5kg) falls on the inclined face of a wedge (inclination 30 degrees & 15kg) from a height of 10m. Assuming all surfaces frictionless and e=1, find the final velocity immediately after the impact.
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We have a collision problem involving a 5 kilogram sphere falling from a height of 10 meters onto a 15 kilogram wedge. The wedge has an inclined surface at 30 degrees to the horizontal. All surfaces are frictionless, and the coefficient of restitution is 1, indicating a perfectly elastic collision. We need to find the final velocities of both objects immediately after the impact.
To find the sphere's velocity just before impact, we apply conservation of energy. Initially, the sphere has potential energy mgh and zero kinetic energy. Just before impact, all potential energy converts to kinetic energy. Setting mgh equal to one-half mv squared, we solve for v equals square root of 2gh. Substituting the values: v equals square root of 2 times 10 times 10, which gives us 14.14 meters per second vertically downward.
Now we analyze the collision geometry by decomposing the sphere's velocity into components relative to the inclined surface. The sphere approaches with 14.14 meters per second vertically downward. The normal to the 30-degree incline makes a 60-degree angle with the vertical. The normal component is v times cosine 60 degrees, which equals 7.07 meters per second. The tangential component is v times sine 60 degrees, which equals 12.25 meters per second. Only the normal component affects the collision, while the tangential component remains unchanged due to frictionless surfaces.
Now we apply the conservation laws to solve for the post-collision velocities. Conservation of momentum in the normal direction gives us: 5 times 7.07 equals 5 times v1n prime plus 15 times v2n prime, which simplifies to 35.35 equals 5 v1n prime plus 15 v2n prime. The coefficient of restitution equation with e equals 1 gives us: v2n prime minus v1n prime equals 7.07. We now have two equations with two unknowns that we can solve simultaneously.
Now we solve the system of equations step by step. From the restitution equation, we express v2n prime as v1n prime plus 7.07. Substituting this into the momentum equation: 35.35 equals 5 v1n prime plus 15 times the quantity v1n prime plus 7.07. Expanding: 35.35 equals 20 v1n prime plus 106.05. Solving for v1n prime: negative 70.7 equals 20 v1n prime, so v1n prime equals negative 3.54 meters per second. The negative sign indicates the sphere bounces back. Therefore, v2n prime equals 3.54 meters per second.