Answer:
a. V1 = -0.227m/s (negative sign denotes that the direction is towards left)
b. V2 = 0.453m/s
Explanation:
Given
M1 = Mass of Block 1= 20g --- Convert to Kilograms
M1 = 0.02kg
U1 = Initial Velocity of Block 1 = 0.68m/s
M2 = Mass of Block 2 = 40g, --- Convert to Kilograms
M2 = 0.04kg
U2 = Initial Velocity of Block 2 = 0m/s
V1 = Final Velocity of Block 1
V2 = Final Velocity of Block 2
a.
Using Conversation of Linear Momentum
M1U1 + M2U2 = M1V1 + M2V2 --- (U2 = 0)
So,
M1U1 = M1V1 + M2V2
M1U1 - M1V1 = M2V2
M1(U1 - V1) = M2V2 ----- (1)
Since kinetic energy is conserved in elastic collision:
½M1U1² + ½M2U2² = ½M1V1² + ½M2V2² ---- U2 = 0
½M1U1² = ½M1V1² + ½M2V2²
M1U1² = M1V1² + M2V2²
M1U1² - M1V1² = M2V2²
M1(U1² - V1²) = M2V2² ----- (2)
Divide (2) by (1)
M1(U1² - V1²)/M1(U1 - V1) = M2V2²/M2V2
(U1² - V1²)/(U1-V1) = V2
(U1 - V1)(U1 + V1)/(U1-V1) = V2
U1 + V1 = V2 -----(3)
Substitute U1 + V1 for V2 in (1)
M1(U1 - V1) = M2(U1 + V1)
Substitute each values
0.02(0.68 - V1) = 0.04(0.68 + V1)
0.0136 - 0.02V1 = 0.0272 + 0.04V1
0.0136 - 0.0272 = 0.04V1 + 0.02V1
−0.0136 = 0.06V1
V1 = -0.0136/0.06
V1 = −0.22666666666666
V1 = -0.227m/s
b.
From (3), U1 + V1 = V2
V1 = V2 - U1 --- substitute in (1)
M1(U1 - V2 + U1) = M2V2
M1(2U1 - V2) = M2V2
2M1U1 - M1V2 = M2V2
2M1U1 = M2V2 + M1V2
2M1U1 = (M2 + M1)V2
V2 = 2M1U1/(M2 + M1)
V2 = (2*0.02*0.68)/(0.04+0.02)
V2 = 0.453333333333333
V2 = 0.453m/s
Answer:
A = -0.0227 m.s–¹
B = 1.36 m.s–¹
Explanation:
Since it is given that:
mass of the smaller block m¹= 0.02
and the velocity of the smaller block u¹ = 0.68m.s–¹
then mass of the bigger block is m = 0.04
with its velocity represented as u = 0 m.s–¹
Hence,
a)
Using conservation of linear momentum:
m¹* u¹ + m * u = m¹* v¹ + m * v
where:
v¹ = final velocity of the smaller block
v = final velocity of the bigger block
m¹* u¹ = m¹*v¹ + m * v
m¹(u¹* v¹)= ........................(1)
Based on kinetic energy being conserved in elastic collision:
½m¹* u¹^² = ½m¹* v¹^² + ½ m * v²
= m¹ (u¹^² - v¹^²) = m * v²
= m1 (u¹ - v¹) (u¹ + v¹^²)
divide the above equation by eq. (1)
v = u¹ + v¹ .............................(2)
now we substitute the value of v from eq. (2) in eq. (1)
m¹ ( u¹ - v¹) = m(u¹ - v¹)
m¹ + m/ m¹ - m = u¹/v¹
= 0.02+0.04 ÷ 0.02-0.04 = 0.68/v1
Therefore v¹ = -0.227 m.s–¹
(negative sign denotes that the direction is towards left)
b)
now we substitute the value of v' from eq. (2) in eq. (1)
m¹ (u¹ - v + u¹) m * v
= 2m¹ * u¹ = (m - m¹) v
= 2 * 0.02 * 0.68 = (0.04 - 0.02) * v
v = 0.0272 ÷ 0.02
v = 1.36m.s–¹
