A 6.0 kg wooden block is suspended by a 1.0 m long wire of negligible mass. A 0.

Question

A 6.0 kg wooden block is suspended by a 1.0 m long wire of negligible mass. A 0.0012 kg bullet strikes the wooden block and embeds in it. After the impact, the bullet and the block together swing along a circular path.

a) If the speed of the bullet and block at the top of the circle is 5.0 m/s, what is the tension in the wire?
b) What is the kinetic energy of the bullet and block at the top of the circle?
c) What is the potential energy of the bullet and block at the top of the circle? *Take the reference level to be the bottom of the circle*
d) What is the kinetic energy of the bullet and block just after the collision?
e) What is the initial speed of the bullet before the collision?

*I understand this is a long problem, but this is my last question to use and I desperately need some assistance on this type of problem. I will rate and thumbs up!*

Explanation / Answer

a.) At the top of the circle, the centrifugal force acts radially outwards, that is vertically upwards. And the gravitational force (Mg) acts vertically downwards. And the tension (T) pulls the mass downwards.

As we know the centrifugal acceleration is given by v2 /R , where v is the velocity and R is the radius of the circular path. So, centrifugal force = M v2 /R

Since there is no vertical acceleration, the net vertical force should be zero. Which means

Net vertical forces = Net horizontal forces

M v2/R = Mg + T

T = ( Mv2 /R - Mg )

putting v = 5m/s , R =1 m and M = 6.0012 kg (combined mass of bullet and the block) , we get

T = ( 6.0012 x 52 / 1 - 6.0012 x 9.81 ) = 91.158228 N

b.) KE = 0.5 M v2 = 0.5 x 6.0012 x 52 = 75.015 J

c.) PE = Mgh = 6.0012 x 9.81 x 2 (since height attained is 2 R = 2x1 = 2 m)

So, PE = 6.0012 x 9.81 x 2 = 117.743544 J

d.) According to the law of conservation of energy, the total energy remains constant

So, Total energy at the top of the circle = Total energy at the bottom of the circle

KE at the top + PE at the top = KE at the bottom + PE at the bottom

75.015 + 117.743544 = KE at the bottom + 0

KE at the bottom = 192.758544 J

e.) For this to be the KE of the combined mass just after the collision, let the velocity be V

So, KE = 0.5 MV2

192.758544 = 0.5 x 6.0012 x V2

V = 8.014985964 m/s

So, the is the velocity of the combined mass just after the collision.

According to the law of conservation of linear momentum, the net momentum before collision needs to be equal to the net momentum after collision.

So, mvo   = MV

where m is the mass of bullet (0.0012 kg ) alone and vo is the initial velocity of the bullet before collision and M is the combined mass and V is the velocity of the combined mass just after the collision.

So, mvo   = MV

0.0012 x vo = 6.0012 x 8.014985964

vo = 40082.9448 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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