Question

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.40 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.50 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use + to mean to the right, - to mean to the left.)(a) What is the velocity of the first block after the collision?(b) What is the velocity of the second block after the collision?Remember that the blocks cannot pass through each other!

Answer 1

Answer:

Explanation:

m = 0.40 kg

u = 2.4 m/s

m' = 0.50 kg

u' = 0 m/s

Let v be the velocity of first block and v; be the velocity of second block after collision.

Use conservation of momentum

m x u + m' x u' = m x v + m' x v'

0.40 x 2.4 + 0 = 0.4 x v + 0.5 x v'

0.96 = 0.4 v + 0.5 v'

9.6 = 4 v + 5v' ..... (1)

According to the conservation of energy

0.5 x m x u² + 1.2 = 0.5 x m x v² + 0.5 x m' x v'²

0.5 x 0.4 x 2.4 x 2.4 + 1.2 = 0.5 x 0.4 x v² + 0.5 x 0.5 x v'²

4.704 = 0.4v² + 0.5v'²

47.04 = 4v² + 5v'²   .... (2)

From equation (1) , v = (9.6 - 5v') / 4 = 2.4 - 1.25 v' , Put in equation (2), we get

47.04 = 4 (2.4 - 1.25v')² + 5v'²

47.04 = 4 (5.76 + 1.56 v'² - 6 v') + 5v'²

47.04 = 23.04 + 6.24v'² - 24v' + 5v'²

11.24v'² - 24v' - 24 = 0

$v'=\frac{24\pm \sqrt{24^{2}+4\times 24\times 11.24}}{2\times 11.24}$

By solving, v' = 2.88 m/s or - 0.74 m/s

So, v = - 1.2 m/s or 3.33 m/s

Answer 2

Answer:

The velocity of the first block is 1.15m/s while of the second block 2.56m/s.

Explanation:

Momentum is only conserved in an isolated system, and because this problem requires us to find the value of the two variables, we need two equations; therefore, to conserved momentum the energy must be released in to the system only after the collision has occurred.

Therefore, from conservation of momentum

$m_1v_1= m_1v_{1f}+m_2v_{2f}$

$0.4(2.4)= 0.4v_{1f}+0.5v_{2f}$

$\boxed{\:0.96= 0.4v_{1f}+0.5v_{2f}}$

and from conservation of energy

$\frac{1}{2}m_1v_1^2+1.2 = \frac{1}{2} m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$

$m_1v_1^2+2.4 = m_1v_{1f}^2+m_2v_{2f}^2}$

$\boxed{\:3.804= 0.4v_{1f}^2+0.5v_{2f}^2.}$

Thus, we have two equations and two unknowns

$(1). \: 0.4v_{1f}+0.5v_{2f}=0.96$

${(2). \: 0.4v_{1f}^2+0.5v_{2f}^2 =3.804$

which has solutions

$(v_{1f}, v_{2f}) = (1.15,2.56)$

and

$(v_{1f}, v_{2f}) = (1.15,-2.56)$

Since the blocks cannot pass through each other, the 0.5kg block cannot have $v_{2f} = -2.56m/s$  (moves to the left) while the 0.4 kg block has $v_{1f} = 1.15m/s$ (moves to the right); therefore, we take the first solution for the velocities:

$(v_{1f}, v_{2f}) = (1.15,2.56)$.

Thus , the velocity of the first block is 1.15m/s while of the second block 2.56m/s.