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 "stone" is $e=0.8$ determine the time required just after collision for $B$ to slide off the runway. This requires the horizontal component of displacement to be $3 \mathrm{ft}$.

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.

00:06 Curling stone a slides over ice and strikes curling stone b, at which point they both move off.

00:14 We want to calculate the time it'll take for curling stone b to move off the runway after the collision.

00:24 So we've drawn our x and y axes as follows.

00:29 Now using these axes, we'll first look at the x, axis and use the conservation of momentum.

00:37 So the conservation of momentum tells us that the momentum before the collision must equal the total momentum after the collision.

00:46 Mv1 must equal to mv2.

00:49 So if we take directions along our x -axis to the left to be positive and we apply the conservation of momentum, we know that the momentum of b initially is zero since b is at rest, plus the momentum of a, which is its weight, over g which gives us its mass 47 over 32 .2 feet per square second times its velocity in the x direction 8 cosine 30 degrees is equal to the momentum after the collision which is 47 over 32 .2 times b's velocity in the x direction vb x2 plus the mass of a times a's velocity in the x direction after the collision.

01:43 So subscript two denotes post -collision.

01:47 So we have here an equation with two unknowns, vv x2 and v -a -x2.

01:53 Now if we choose this direction once more, but now look at the coefficient of restitution, the coefficient of restitution e is given to be 0 .8.

02:06 And by definition this is equal to vb x2 minus the a x2 after the collision over 8 cosine 30 degrees minus zero so now we have another equation with the same two unknowns vb x2 and v a x2 if we solve these two equations we can find these two unknowns so we solve these two unknowns so we them simultaneously and we get vax2, the x component of a's velocity after the collision is 0 .6928 feet per second.

03:03 The x component of b's velocity after the collision vbx2 is 6 .2335 feet per second...

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