# Abaseball of mass m1 = 0.45 kg is thrown at another ball hanging from the ceiling by a length of string

###### Question:

## Answers

0.947 rad or 54.27 degrees

Explanation:

Suppose the collision is elastic, meaning the momentum is preserved.

Before the collision [tex]v_2 = 0 m/s[/tex] and the second ball has no momentum

Right after the collision [tex]v_1 = 0 m/s[/tex] and the first bast ball has no momentum.

Therefore momentum of the first baseball has been transferred to the 2nd ball

[tex]m_1v_1 = m_2v_2[/tex]

[tex]4.5 * 0.45 = 0.61v_2[/tex]

[tex]v_2 = \frac{4.5 * 0.45}{0.61} = 3.32 m/s[/tex]

After the collision, the second ball would gained kinetic energy, which then would be transferred to potential energy once it reaches its highest point:

By the law of energy conservation:

By law of energy conservation:

[tex]E_k = E_p[/tex]

[tex]\frac{mv^2}{2} = mgy[/tex]

[tex]v^2 = 2gy[/tex]

[tex]y = \frac{v^2}{2g} = \frac{3.32^2}{2*9.81} = 0.562 m[/tex]

So at its highest point, the ball is 0.562 m from the lowest point. Since the ball is hanging on a 1.35 m string, we can calculate the vertical distance from there to the swinging point:

1.35 - 0.562 = 0.788 m

Finally, the angle that string makes with the vertical at the highest point is

[tex]\alpha = cos^{-1}(\frac{0.788}{1.35}) = 0.947 rad = 54.27^o[/tex]

50 minutes

Step-by-step explanation:

C(m) =0.1m+1.1 = 6.10 ... - 1.1 both sides

0.1 m = 5.0

m = 50

The angle is 16.86 °

Explanation:

Step 1: Given Data

mass of the baseball 1 = 0.28 kg

lengt of string L = 1.35m

mass of ball 2 = 0.92 kg

horizontal velocity = 3.5 m/s

After the collision the first baseball falls straight down (no horizontal velocity).

Step 2: Calculate the velocity

The linear moment before the shock is equal to the linear moment after the shock:

m 1 *u 1 + 0 = 0 + m2*v

v= u1 * m1/m2 = 3.5m/S * ( 0.28/0.92)

v = 1.065 m/s

Step 3: Calculate the height

The kinetic energy of ball 2, immediately after the shock, is equal to the final gravitational potential energy

Kinitial = Ug, final

1/2*m2*v² = ms * g*h

h= v²/2g

h= 1.065²/(2*9.81)

h = 0.058 meter

Step 4: Calculate the maximum angle the rope makes with the vertical is:

cos(∅) = (L-h)/L

∅ = cos^-1 ((1.35-0.058)/1.35)

∅ = cos^-1(0.957)

∅ =16.86°

The angle is 16.86 °

The angle is approximately 50°

Explanation:

We know,

h = L – L * cos θ

= 1.55 – 1.55 (cos θ )

To determine the maximum height, we use conservation potential and kinetic energy. As 2nd ball rises to its maximum height, the increase of its potential energy is equal to the decrease of its kinetic energy. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.

Potential energy, PE = mgh = m X 9.8 X h

Kinetic energy, KE = [tex]\frac{1}{2} mv^2[/tex]

Set PE equal to KE and solve for h.

h = v²/19.6

To determine the initial kinetic energy of the second ball, we need to the velocity of the 2nd ball immediately after the collision. To do this we need to use conservation of momentum. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.

For the 1st ball, horizontal momentum = 0.26 X 2.5 = 0.65

Since this ball falls straight down after the collision, its final horizontal momentum is 0.

For the 2nd ball, horizontal momentum = 0.61 X vf

0.64 * vf = 0.65

vf = 0.65/.0.64

vf = 0.02m/s

h = (0.65/0.64)²/19.6

h = 0.00002m

0.00002 = 1.45 – 1.45 * cos θ

Subtract 1.55 from both sides.

0.00002 – 1.45 = -1.45 * cos θ

θ = 50°

the approximately angle = 34⁰

Explanation:

given

[tex]m_{1} = 0.35 kg[/tex]

length of string L = 1.35 m

[tex]m_{2} = 0.93 kg[/tex]

initial horizontal velocity

[tex]v_{1} = 3.5 m/s[/tex]

suppose ball will rise up to a height h as

h = L – L cos θ

h = 1.35 – 1.35 cos θ

1.05 cos θ = 1.35 – h

at the maximum height, we use conservation of kinetic and potential energy. As the 2nd ball rises to its maximum height, its kinetic energy will convert into potential energy.

KE = PE

[tex]PE = KE\\mgh = \frac{1}{2} mv^2\\h = \frac{v^2}{2g}[/tex]

[tex]h =\frac{v^2}{2g} = \frac{v^2}{19.6}[/tex]

we kinetic energy of the second ball after the collision. To determine the 2nd ball velocity after the collision,

use conservation of momentum.

Initial momentum of 1st ball = 0.35 × 3.5 = 1.225 ..........(i)

Since the baseball has no horizontal velocity after the collision, all of its momentum is transferred to the 2nd ball.

momentum for second ball = mv = 0.93 v ......(ii)

equate (i) and (ii) we get

0.93 v = 1.225

v = 1.32 m/s .........(iii)

use this value to find height

[tex]h = (1.32)^2/19.6[/tex]

This is approximately 0.088 meter

cos θ = (1 – 0.08)/1.35

cos θ = 0.9407

the approximately angle = 34⁰

the final velocity after collision of the second ball is 1.065m/s

Explanation:

this question is about momentum changes which are equal in magnitude and opposite in direction.

the law of momentum conservation stated that for a collision occurring between object1 and object2 in an isolated system, the total momentum of the two object before collision is equal to the total momentum after collision, (www.physicsclassroom.com).

therefore. m1*(Δv1) = -m2* (Δv2)

m1= 0.28kg, v₁₁ = 3.5m/s, v₁₂ =0

m2= 0.92kg v₂₁ = 0m/s, v₂₂ =?

the final velocity of the second ball after collision is to be determined:

applying the formula: m1*(Δv1) = -m2* (Δv2)

= 0.28* (3.5 - 0) = -0.92 (0 - v₂₂)

= 0.98 = 0.92v₂₂

v₂₂ = 0.98/0.92 = 1.065m/s

the final velocity after collision of the second ball is 1.065m/s