Block A is released from rest and slides down the...

Key Concepts

Conservation of Linear Momentum

This principle states that in an isolated system where external forces are negligible during the short time of collision, the total linear momentum remains constant. In collisions, even when energy transformations occur (such as loss in inelastic impacts), the momentum before and after the collision must balance. This concept is crucial when solving for the velocities of interacting bodies post-collision.

Coefficient of Restitution

The coefficient of restitution (e) measures the elasticity of a collision; it is defined as the ratio of the relative speed of separation to the relative speed of approach between two colliding bodies. A value of e = 1 indicates a perfectly elastic collision with no kinetic energy loss in the normal component, while e = 0 signifies a perfectly inelastic collision where the bodies may move together post-collision. This parameter directly influences the post-collision velocities.

Relative Velocity in Collisions

In collision analysis, the relative velocity of the two bodies before impact, and its transformation after impact, is key to determining their individual speeds. The relationship between the velocities, as mediated by the coefficient of restitution, describes how the velocity difference between the two bodies reverses or diminishes following the collision. This concept is essential in setting up the equations that link momentum conservation with the energy loss or retention characterized by the coefficient of restitution.

Transcript

00:01 This question, we have mass of block a, m .a equals to 10 kg, and mass of block b, mb equals to 30 kg, and b can roll freely on the ground.

00:13 So the block a is released from the rest and slides down on the freshness surface of b until it hits a bumper on the right end of b.

00:22 So now, when a is sliding down, so we can write from the momentum conservation, that is ma, v, a, x plus mb, b, vb x equals to 0 okay this is our equation number 1 similarly when a is impacted to the b we can apply moment of conservation so we can write m a v a x dash plus m b b b b b x dash equals to 0 this is our equation number 2 now applying work energy principle so we can write that kinetic energy t1 plus potential energy v1 equals to kinetic energy t2 plus potential energy energy v2 this is initial kinetic energy and this is finite okay so we can see that t1 equals to 0 block because block a a1 is rest and t2 is equal to 1 by 2 m a v a square plus 1 by 2 m b b b b square okay so and v2 is equal to 0 now we can write also for the v1 which will be equals to vb which a plus a and b so which is equal to m a g h plus zero which comes out to be m a g h okay so now substituting all the values in this equation three we get zero plus m a g h this is equals to one by two m a v a square plus one by two m b b square plus zero okay now since v a is equals to v a x so we can write from equation one this equation that is vb x will be equals to minus of m a divided by m b multiplied by v a x okay so now substituting these values in this equation we get m a g h this is equals to 1 by 2 m a v a square plus 1 by to m b and b bx or simply vb this will be equals to minus of m a divided m b multiplied by m b multiplied by v a x whole square okay so now after solving for v a the velocity of block a before impact comes out to be v a equals to under root of 2 g h divided by 1 plus m a divided by m b so now substituting this equation in the uh substituting this equation in the equation number 1 we get vb equals to minus of m a divided by mb multiplied by v a or v a x which is equal to under root of 2g h divided by 1 plus m a divided by mb okay so now we have all the values and h is equal to 0 .2 meter so now substituting all the values we get v a equals to v b sorry v a x this comes out to be 1 .71 w b this comes out to be 1 .7 1 .7 5 meter per second after substituting values in this equation.

03:41 So now the relation between relative velocities of a and b before and after impact from the equation is given by vvx minus vax dash.

03:55 This becomes equals to e of vax minus vbx, okay, where e is the coefficient of restitution.

04:03 So substituting value of vbx, we get vbx -bx -v -b -x -v -b -s -d -s -m -a.

04:08 V a x dash and vbx dash this can also be written in the form of vax as minus of m a divided by mb multiplied by v a x dash okay this equals to e times of v a x minus v b b b b is minus m a divided by m b multiplied by v a x okay so now from here after solving further we get v a x this equals to minus times of e, vax.

04:45 Now, substituting value of vax, which is equals to this value...

Please give Ace some feedback