A block A [2kg], push against a spring by x = 0.1 m. A ball B [1 kg], hanged on

Question

A block A [2kg], push against a spring by x = 0.1 m. A ball B [1 kg], hanged on string at rest, the length of the string L = 1 m. A is released from a rest (i.e., V_ = 0) and it is sliding down along ramp with an angle of 20 degree and sliding length of d = 1 s m, then an hits the ball B, the B is moving up until it reaches the position of the angle alpha = 30 degree. Other known parameters are: spring constant k = 800 N/m friction coefficient mu = 0.2 coefficient of restitution e = 0.8. Please determine: 1) speed of A just before it hit the ban B (after travels x+d distance.) 2) Speed of ball B immediately after the impact. 3) Speed of B when it reaches the position with angle of alpha = 30 degree 4) Tension force in the string at the position of the angle alpha = 30 degree 5) Impulse on the ball B in the impact.

Explanation / Answer

1)

Let the speed of A just before the impact be v.

Work doon the A is;

W = (1/2)k(x)2 + mgsin200(x+d) - µmgcos200(x+d)

or, W = (0.5)(800 N/m)(0.1 m)2 + mg(x+d)[sin200 - µcos200]

or, W = (0.5)(800 N/m)(0.1 m)2 + (2 kg)(9.8 m/s2)(0.1 m + 1.5 m)[sin200 - µcos200]

or, W = 8.83 J

Work done is equal to change in kinetic energy, so,

W = (1/2)mv2

or, (1/2)(2 kg)v2 = 8.83 J

or, v = 3 m/s, is the speed of A before it hits B.

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2)

A will impart horizontal velocity to B. Let the speed of B immediately after impact be vb and the horizontal velocity of A be va. Coefficient of restitution is e = 0.8

Conservation of linear momentum along horizontal direction says,

mavcos200 = mava + mbvb

or, 2vcos200 = 2va + vb -------- (1)

Also, vb - va = evcos200 = 0.8vcos200 ------- (2)

So from equations (1) and (2), we get,

2vcos200 = 2(vb - 0.8vcos200) + vb

or, 3.6vcos200 = 3vb

or, 1.2(3m/s)cos200 = vb

or, vb = 3.38 m/s, is the speed of B immediately after the impact.

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3)

After the impact the ball moves up till an angle 300

So speed of B at this position would be zero.

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4)

When B is at = 300

T - mgcos300 = 0, since speed is zero at this instant,

So, T = mgcos300 = (2kg)(9.8 m/s2)cos300

or, T = 17 N, is the tension in the string at this position.

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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