The "stone" A used in the sport of curling slides over the...

The "stone" $A$ used in the sport of curling slides over the ice track and strikes another "stone" $B$ as shown. If each "stone" is smooth and has a weight of $47 \mathrm{lb}$, and the coefficient of restitution between the "stones" is $e=0.8$ determine their speeds just after collision. Initially $A$ has a velocity of $8 \mathrm{ft} / \mathrm{s}$ and $B$ is at rest. Neglect friction.

Key Concepts

Conservation of Momentum

This principle states that in a closed system with no external forces the total momentum remains constant before and after a collision. It is fundamental for analyzing collision problems, as it provides an equation that relates the masses and velocities of objects involved in the collision, allowing one to solve for unknown final velocities.

Coefficient of Restitution

The coefficient of restitution is a measure of the elasticity of a collision, defined as the ratio of the relative speed after collision to the relative speed before collision along the line of impact. It quantifies how much kinetic energy is conserved in a collision, with a value of 1 representing a perfectly elastic collision and lower values indicating inelastic behavior.

Sequential Collision Analysis

Sequential collision analysis involves examining collisions that occur one after the other, where the outcome of one collision serves as the initial condition for the next. It requires applying conservation of momentum and the coefficient of restitution for each individual collision in a step-by-step manner to determine the overall system behavior.

Transcript

00:01 In this problem, we have two curling stones a and b.

00:05 Curlingstone a strikes b and after the collision, we need to calculate the speeds at which they both move off.

00:14 So we'll align our x and y axes as we have in the diagram.

00:21 And we'll first look at the line of impact along the x axis and use the conservation of momentum.

00:30 So the conservation of momentum tells us that the momentum before the collision for both stones must equal the total momentum after the collision, so momentum is conserved.

00:44 Now we'll take everything to the left along the x -axis as positive.

00:55 So initially, b has no momentum, so that's zero.

01:00 And stone a has momentum as follows.

01:05 Its mass is 47 pounds divided by g 32 .2 feet a square second.

01:15 Its velocity along the x -axis is 8 times the cosine of 30 degrees and this must equal to the momentum of the system after the collision.

01:27 So the mass of b which is again 47 over 32 .2 times its velocity in the x direction after the collision vb x2 plus the momentum of a after the collision 47 over 32 .2 times vax 2, subscript 2 denotes after the collision.

01:57 So essentially we have an equation here with two are known.

02:00 Vbx2 and vax2.

02:03 So we need to set up another equation and we'll do that by using the same direction but using the coefficient of restitution in e.

02:12 So we know e is equal to 0 .8, and this is equal to, by definition, vb x2 minus v .a x2 over the difference in initial velocities, 8, cosine of 30 degrees, minus for b is 0...

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