TS EAMCET · Physics · Center of Mass Momentum and Collision
A bullet of mass \(m_1\) is moving with speed \(v_0\) hits a sand bag of mass \(m_2\). If the speed of the bullet after passing the sand bag is \(\frac{v_0}{3}\), then the height \(h\) upto which the bag rises is (assume, \(g=\) acceleration due to gravity)
- A \(h=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2\)
- B \(h=\frac{2 m_1 v_0}{3 m_2}\)
- C \(h=\frac{1}{2 g}\)
- D \(h=\left(\frac{2 m_1 v_0}{3 m_2}\right)^2\)
Answer & Solution
Correct Answer
(A) \(h=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2\)
Step-by-step Solution
Detailed explanation
The given situation is as shown in figure, As, momentum is conserved in collision, \(\therefore \quad p_i=p_f\) \(\begin{array}{ll}\Rightarrow & m_1 v_0=m_2 v_2+m_1 \frac{v_0}{3} \\ \Rightarrow & v_2=\frac{2 m_1 v_0}{3 m_2}\end{array}\) Given that, the bag rises upto height…
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