Question

The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision between the blocks is perfectly elastic. The mass m2 then compresses the spring an amount x.

True, False, greater than, equal to, less than :

1. Immediately after colliding with m2, the mass m1 stops and has zero velocity.
2. Immediately after the collision, the momentum of m2 is ....... the initial momentum of m1.
3. The maximum energy stored in the spring is ...... the initial energy of m1.
4. Immediately after the collision, the energy of m2 is ...... the initial energy of m1.

Answer 1

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

• As the collision is perfectly elastic, kinetic energy must be conserved.
• The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

        $v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}$

• As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

• As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)    

• The maximum energy stored in the in the spring is given by the following expression:

       $U =\frac{1}{2} *k * A^{2}$

• where A = maximum compression of the spring.

• This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
• When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
• Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
• As total kinetic energy must be conserved, the following condition must be met:

       $KE_{10} = KE_{1f} + KE_{2f}$

• So, it is clear that KE₂f  < KE₁₀
• Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

• As explained above, if total kinetic energy must be conserved:

        $KE_{10} = KE_{1f} + KE_{2f}$

• So as kinetic energy is always positive, KEf₂ < KE₁₀.

Answer 2

(1) Immediately after colliding with m2, the mass m1 stops and has zero velocity [False].

(2) Immediately after the collision, the momentum of m2 is greater than the initial momentum of m1.

(3) The maximum energy stored in the spring is is less than the initial energy of m1 due to energy lost to friction.

(4) Immediately after the collision, the energy of m2 is equal the initial energy of m1.

The given parameters:

• first mass, = m₁
• second mass, = m₂
• initial velocity of the first mass, = v₀
• initial velocity of the second mas = 0
• extension of the spring, = x

According the principle of conservation of linear momentum, the total initial momentum is equal to the total final momentum.

$m_1 v_0_1 \ + \ m_2 v_0_2 = m_1 v_f_1 + m_2 v_f_2\\\\m_1v_0 + m_2(0) = m_1(-v_f_1) \ + \ m_2v_f_2\\\\m_1v_0 = -m_1v_f_1 + m_2v_f_2\\\\m_1v_0 + m_1v_f_1 = m_2v_f_2$

After the collision, the final velocity of the block m₁ is $v_f_1$.

Thus, immediately after colliding with m2, the mass m1 stops and has zero velocity [False].

Immediately after the collision, the momentum of m2 is greater than the initial momentum of m1.

$m_1v_0 + m_1v_f_1 = m_2v_f_2$

The maximum energy stored in the spring is calculated as;

$U_x = \frac{1}{2} k A^2 = P.E + K.E \ - \mu Fd$

The maximum energy stored in the spring is is less than the initial energy of m1 due to energy lost to friction.

In elastic collision, kinetic energy is conserved,

Thus, immediately after the collision, the energy of m2 is equal the initial energy of m1.

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