Question

a 1.2kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of .05m/s. A the origin 0, the bomb explodes into two fragments. Fragment 1 has a mass of .40kg and a speed of .90m/s along the negative y-axis. Find the magnitude and direction of the velocity of fragment 2.

Answer 1

Answer:

$80.54^{\circ}$

$0.46\ \text{m/s}$

Explanation:

m = Mass of initial piece = 1.2 kg

$u_x$ = Velocity of toy in x direction = 0.05 m/s

$u_y$ = Velocity of toy in y direction = 0

$v_{1x}$ = Velocity of fragment 1 in x direction = 0

$v_{1y}$ = Velocity of fragment 1 in y direction = -0.9 m/s

$v_{2x}$ = Velocity of fragment 2 in x direction

$v_{2y}$ = Velocity of fragment 2 in y direction

$m_1$ = Mass of fragment 1 = 0.4 kg

$m_2$ = Mass of fragment 2 = 1.2-0.4 = 0.8 kg

Applying conservation of momentum in x axis

$mu_x=m_1v_{1x}+m_2v_{2x}\\\Rightarrow 1.2\times 0.05=0.4\times 0+0.8\times v_2\cos\theta\\\Rightarrow v_2\cos\theta=\dfrac{1.2\times 0.05}{0.8}\\\Rightarrow v_2\cos\theta=0.075\ \text{m/s}$

Applying conservation of momentum in y axis

$mu_{y}=m_1v_{1y}+m_2v_{2y}\\\Rightarrow 0=0.4\times -0.9+0.8\times v_{2y}\\\Rightarrow v_{2y}=\dfrac{0.4\times 0.9}{0.8}\\\Rightarrow v_2\sin\theta=0.45\ \text{m/s}$

From the above two final equations we get

$\dfrac{v_2\sin\theta}{v_2\cos\theta}=\dfrac{0.45}{0.075}\\\Rightarrow \theta=\tan^{-1}\dfrac{0.45}{0.075}\\\Rightarrow \theta=80.54^{\circ}$

The direction of the fragment 2 is $80.54^{\circ}$

$v_2=\dfrac{0.075}{\cos\theta}\\\Rightarrow v_2=\dfrac{0.075}{\cos80.54^{\circ}}\\\Rightarrow v_2=0.46\ \text{m/s}$

The velocity of fragment 2 is $0.46\ \text{m/s}$

Answer 2

The magnitude and direction of the velocity of fragment 2 is 0.46 m/s at 80.54⁰.

The given parameters;

• mass of the spring, m = 1.2 kg
• speed of the spring, v = 0.05 m/s
• mass of the first fragment, m₁ = 0.4 kg
• initial speed of the first fragment, u₁ = 0.9 m/s
• mass of the second fragment, m₂ = 0.8 kg

Apply the principle of conservation of linear momentum, to determine the speed of the second fragment;

in x-direction:

$mucos\ \theta = m_1 v_1 cos\theta + m_2 v_2cos\theta\\\\1.2 \times 0.05 \times cos (0) = 0 + 0.8v_2cos \theta\\\\v_2cos \theta = \frac{0.06}{0.8} \\\\v_2cos \theta = 0.075 \ m/s$

In y-direction:

$1.2(0) = -0.4(0.9)+ 0.8(v_2sin\theta)\\\\v_2 sin\theta = \frac{0.36}{0.8} \\\\v_2 sin\theta = 0.45$

solve the first and second equation together;

$\frac{v_2 sin\theta }{v_2cos \theta} = \frac{0.45}{0.075} \\\\tan\theta = 6 \\\\\theta = tan^{-1}(6)\\\\\theta = 80.54 \ ^0$

The value of the velocity is calculated as;

$v_2 sin\theta = 0.45\\\\v_2 = \frac{0.45}{sin\theta} \\\\v_2 = \frac{0.45}{sin(80.54)} \\\\v_2 = 0.46 \ m/s$

Thus, the magnitude and direction of the velocity of fragment 2 is 0.46 m/s at 80.54⁰.

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