A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block has a head-

Question

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision.

(a) What is the speed of the 20.0 gram block after the collision?
(b) What is the speed of the 40.0 gram block after the collision?

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Verity 3 years 2021-08-24T05:44:09+00:00 2 Answers 81 views 0

Answers ( )

  1. ngocdiep
    0
    2021-08-24T05:45:37+00:00
    Reply

    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

  2. Doris
    0
    2021-08-24T05:46:08+00:00
    Reply

    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

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