Speed of a bullet

Let the mass of the block be M, the mass of the bullet be m, the velocity of the bullet be v, the acceleration due to gravity be g, the height at which the block rose be h, the length of the string be l, the maximum angle be A (A= 60 degrees by condition), and the velocity of the system block+bullet when the bullet hits the block be u.

Consider the moment when the bullet hits the block. Then we have the initial kinetic energy of the system (bullet+block) equal to $(m+M)v^2/2$ . After the collision we have the potential energy of the system equal to $(m+M)gh$ . According to he Law of Conservation of Energy, these must be equal. So, equating...

$(m+M)u^2/2= (m+M)gh$ Or, $u= \sqrt{2gh}$ .....(1)

Now again consider the momentum before the collision. The total momentum is mv. Now consider the momentum during the collision. The total momentum is (m+M)u (since the whole system bullet+block is moving now, and it has a mass of m+M). According to the Law of Conservation of Momentum, these values must be equal.

Equating...

$(m+M)u= mv$ Or, $v= (m+M)u/m$

But from (1) we know that $u= \sqrt{2gh}$ . Substituting these values, we get $u= (m+M)\sqrt{2gh}/m$ .....(2)

But the value of h is not given. This has to be found out. We know that the angle between the final position of the block and the initial position of the block is 60 degrees. Consider a triangle whose three vertices are the initial position of the block, the final position of the block, and the other endpoint of the string. We know that this triangle is isosceles, since two sides of this triangle represent the length of the string, which is constant. It is also given that the angle between the equal sides of this triangle is 60 degrees. Using geometry, we can prove that the other angles must also be 60 degrees (the other angles must be equal, let they be X degrees. So, X+X+60= 180 -> X= 60). So the triangle must be equilateral. Now consider the line joining the initial and the final position of the block. By our previous arguments this length must be equal to x. Now the angle which the string subtends with the horizontal when the block is at its final position is (90 - 60) degrees= 30 degrees. So, we have $h= l sin(30~degrees)= l/2$ .

Plugging this value into (2), we get $v= (m+M)\sqrt{gl}/m$ .

Here M= 1 kg, m= 10 gram= 1/100 kg, g= 9.8 m/s^2, l= 1 m. Plugging the values, we get the initial muzzle velocity to be 316.18 m/s.

M ( mass of block ) = 1 kg, m ( mass of bullet ) = 0.01 kg, l ( length of string ) = 1m, g = 9.8 ms^-2 . Let u = initial speed of the block, v= speed of the block just after collision.

By Conserving Energy:

Kinetic Energy lost = Potential Energy gained

1/2 * (M+m) * v^2 = (M+m) * g * l(1-cos 60)

v^2 = 2 * g * 1/2

v= 7 * (0.2)^1/2

By Conservation of Momentum:

mu=(M+m)v

0.01 * u = 1.01 * 7 * (0.2)^1/2

u= 707 * (0.2)^1/2

u= 316.18 ms^-1

Since there is no external force acting on (block+ bullet) system, total momentum of the system remains conserved.

Let velocity of bullet be v, and velocity of block just after impact = v' After collision, the bullet gets stuck, which means the effective mass of block now is (1 + 10^-2) kgs.

then, (10^-2)(v) = (1 + 10^-2)(v') or v' = (10^-2) v/ (1+ 10^-2) ...(1)

Now, the block's angular amplitude is pi/3 radians. Length of the string = 1m ; Which means, the block rises up to a height h = 1 - (1 * cos(pi/3) ) = 0.5m

The block stops at that point, which implies that all the kinetic energy of the block acquired just after collision must have been converted into potential energy, hence, (1/2)(1 + 10^-2) (v' ^2) = ( 1 + 10^-2)(9.8)(0.5)

Substituting value of v' from eqn 1 , and solving for v, we get our desired answer.

As the bullet strikes the block, some energy may be lost but momentum is conserved. Let v is the speed of bullet and V be the combined speed of block and bullet. By conservation of momentum mv = (m+M) V where m and M are the masses of bullet and block respectively. After the impact, both the bullet and block swings on the string. If we take block and bullet as our system then we can apply law of conservation of energy as the only forces involved are conservative forces. Therefore, $1/2(m+M)V^2 = (m+M)gh$ Eliminating V from both the equations we get the value of v which is required.

The approach to this problem is to solve by working backwards.

We know the block swings a maximum of 60 degrees with respect to the vertical. We also know the vertical is 1 m long. In order to use the formula for gravitational potential energy ( $PE=mgh)$ we need to figure out the change in height. To solve for the change in height draw a triangle with a 1 m hypotenuse and sixty degree angle from the vertical. By the rules of 30-60-90 triangles the leg on the vertical must be $\frac {1}{2}$ . Thus the change is height is $1-\frac {1}{2}$ .

Thus $PE=\frac {mg}{2}$ . The kinetic energy( $KE=\frac {mv^2}{2}$ ) we do not know, but we know that after the bullet enters the block, mechanical energy is conserved. $KE+PE=$ a constant

At the maximum height the kinetic energy is 0, there is no velocity. And we can set the 0 level of potential energy at the lowest point of the block. Thus our equations become: $\frac {mv_1^2}{2} = \frac {mg}{2}$ Where $v_1$ is the velocity of the block after the bullet has struck it. Solving for v, we get $v = sqrt{g}$ or approximately 3.13. Some of you may be worried the units do not match up, look carefully, can you figure out why? (Be very observant when we calculated the change in height).

Now comes time to apply momentum. Before the bullet hits the block, the bullet is the only object moving. Call its velocity $v_2$ and its mass $m_2$ . And the mass of the block is $m_1$ . After the bullet hits the block, the bullet and the block become one object. Thus we obtain the equations

$v_2m_2 = (m_2 + m_1)v_1$

Solving for: $v_2 = \frac{1.01kg*3.13m/s^2}{0.01kg} = 316m/s$

The block/bullet combination get a total energy once the bullet hits the block. Since they swing to 60 degrees with respect to the vertical, the net gain in potential energy is $\Delta U=(m_b+m_B)gL(1-cos60)$ , where $m_b$ is the mass of the bullet, $m_B$ is the mass of the block, and L is the length of the string. Since energy is conserved, this is the same as the initial kinetic energy right after the bullet hits the block, i.e.

$\frac{1}{2} (m_b+m_B)v^2=\frac{1}{2} (m_b+m_B)gL$ .

The velocity is then $v=\sqrt{gL}=\sqrt{g}$ . Momentum is conserved in the collision of the bullet and block, hence $v_b=(m_b+m_B) v/m_b$ , yielding $v_b=316.18~m/s$ .