Question

A baseball of mass m1 = 0.28 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.35 m. The second ball m2 = 0.92 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 3.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity.)

Answer 1

The angle is 16.86 °

Step-by-step 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 °

Answer 2

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

Step-by-step 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