2) A ball is attached to one end of a wire, the other end being fastened to the

Question

2)

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.7 kg and 2.30 kg, and the length of the wire is 1.27 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Explanation / Answer

conservation of momentum

0.004 * 368 + 1.15 *0 + 1.53 * 0 = 1.154 * 0.510 + (1.53 + 0.004) * v

0.0065 * 352 = 1.201 * 0.644 + 1.5355 * v

v = 0.575 m/s ---answer
velocity of the second block after the bullet imbeds itself

b)
total kinetic energy after the collision
= 0.5* 1.15 * 0.510^2 + 0.5 *(1.53 + 0.004) * 0.986^2 = 0.895 j

total kinetic energy before the collision
= 0.5 *0.004 * 368^2 = 270.8 j

ratio of the total kinetic energy after the collision to that before the collision
= 3.3 x 10^-3

2 )

the ball converts its PE to KE, so has a speed before impact of

1/2 m v^2 = m g L

v = Sqrt[2 g L] = Sqrt[2 x 9.8m/s/s x 1.27m] = 4.98m/s directed toward the right

the collision is elastic, so both momentum and energy are conserved, we have then:

from momentum conservation:

1.7kg x 4.98m/s = 1.7 kg Va + 2.3 Vb where Va, Vb are the speed of the ball, block after collision

energy conservation gives us

1/2 x 1.7kg x (4.98m/s)^2 = 1/2 x 1.7kg x Va^2 + 1/2 x 2.3 kg x Vb^2

we can write the momentum equation as

Vb = (8.46-1.7Va)/2.3 = 3.6 - 0.73Va

substitute this into energy:

u will get Va

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.