Respuesta :
Answer:
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
Answer:
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 °
