A particle of mass 5m moving with speed v explodes and splits into two pieces with masses of 2m and 3m. the lighter piece continues to move in the original direction with speed 5v relative to the heavier piece. What is the actual speed of the lighter piece?
5 m v = 2 m (x) + 3 m (x/5)
5 v = 2 x + 3x/5
25 v = 10 x + 3 x = 13 x
x = (25/13) v
To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum of a system remains constant before and after the explosion.
Before the explosion, the total momentum of the system is given by:
Initial momentum = mass × velocity = (5m) × v = 5mv
After the explosion, the total momentum of the system is:
Final momentum = (mass of first piece) × (velocity of first piece) + (mass of second piece) × (velocity of second piece)
Let's assume the velocity of the first piece (mass 2m) is v1 and the velocity of the second piece (mass 3m) is v2.
Final momentum = (2m) × v1 + (3m) × v2
According to the given information, the velocity of the first (lighter) piece is 5v relative to the second (heavier) piece. So we can write:
v1 = 5v + v2
Now, we can express the total momentum of the system as:
5mv = (2m) × (5v + v2) + (3m) × v2
Simplifying the equation:
5mv = 10mv + 2mv2 + 3mv2
5mv - 10mv = 5mv2 + 3mv2
-5mv = 8mv2
-5 = 8v2
v2 = -5/8
Since speed cannot be negative, we take the absolute value of v2:
v2 = 5/8
Now, let's substitute the value of v2 back into the equation for v1:
v1 = 5v + v2
v1 = 5v + 5/8
v1 = (40v + 5)/8
Therefore, the actual speed of the lighter piece is (40v + 5)/8.
To determine the actual speed of the lighter piece after the explosion, we can start by using the principle of conservation of momentum. In an isolated system, the total momentum before the explosion is equal to the total momentum after the explosion.
Before the explosion:
The initial momentum is given by:
\(p_{\text{initial}} = m_1 \cdot v_{\text{initial}}\),
where \(m_1\) is the mass of the particle (5m) and \(v_{\text{initial}}\) is its initial speed (v).
After the explosion:
The total momentum is given by:
\(p_{\text{final}} = m_2 \cdot v_{\text{light}} + m_3 \cdot v_{\text{heavy}}\),
where \(m_2\) is the mass of the lighter piece (2m), \(m_3\) is the mass of the heavier piece (3m), \(v_{\text{light}}\) is the speed of the lighter piece, and \(v_{\text{heavy}}\) is the speed of the heavier piece.
According to the conservation of momentum:
\(p_{\text{initial}} = p_{\text{final}}\).
Substituting the given values:
\(m_1 \cdot v_{\text{initial}} = m_2 \cdot v_{\text{light}} + m_3 \cdot v_{\text{heavy}}\).
Since the lighter piece continues to move in the original direction with the speed 5v relative to the heavier piece, we can write:
\(v_{\text{light}} = v_{\text{heavy}} + 5v\).
Substituting \(v_{\text{light}}\) in the equation above, we get:
\(m_1 \cdot v_{\text{initial}} = m_2 \cdot (v_{\text{heavy}} + 5v) + m_3 \cdot v_{\text{heavy}}\).
Simplifying the equation further:
\(5mv = 2m \cdot (v_{\text{heavy}} + 5v) + 3m \cdot v_{\text{heavy}}\).
Expanding the equation:
\(5mv = 2mv_{\text{heavy}} + 10mv + 3mv_{\text{heavy}}\).
Combining like terms:
\(5mv = 5mv_{\text{heavy}} + 10mv\).
Simplifying further:
\(0 = 5mv_{\text{heavy}} + 5mv\).
Rearranging the equation:
\(5mv_{\text{heavy}} = -5mv\).
Dividing both sides by 5m:
\(v_{\text{heavy}} = -v\).
Since the lighter piece is moving in the same direction as the initial motion, its speed can be found by substituting \(v_{\text{heavy}}\) into the equation:
\(v_{\text{light}} = v_{\text{heavy}} + 5v\).
Substituting \(v_{\text{heavy}} = -v\) into the equation above:
\(v_{\text{light}} = -v + 5v\).
Simplifying further:
\(v_{\text{light}} = 4v\).
Therefore, the actual speed of the lighter piece after the explosion is 4 times the initial speed (4v).