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Step-by-Step Solution
Step 1: Determine the velocity of the mass just before the explosion
The mass is thrown up with an initial velocity $u = 100\,\text{m/s}$ and is acted upon by gravity ($g = 10\,\text{m/s}^2$ downward). After $t = 5\,\text{s}$, its velocity can be found using the equation:
$v = u - g t = 100 - (10 \times 5) = 100 - 50 = 50\,\text{m/s}$
Thus, just before the explosion, the mass is moving upwards at $50\,\text{m/s}$.
Step 2: Set up the conservation of momentum
Immediately after exploding, the 1 kg mass splits into two parts:
• One part has mass $0.4\,\text{kg}$ and velocity $-25\,\text{m/s}$ (downwards).
• The other part has mass $0.6\,\text{kg}$ and unknown velocity $v'$.
Before the explosion, the total momentum of the mass was:
$P_\text{initial} = (1\,\text{kg}) \times (50\,\text{m/s}) = 50\,\text{kg·m/s}$
After the explosion, the total momentum is:
$P_\text{final} = (0.4\,\text{kg}) \times (-25\,\text{m/s}) + (0.6\,\text{kg}) \times v'$
By the law of conservation of momentum:
$P_\text{initial} = P_\text{final}$
$50 = 0.4 \times (-25) + 0.6 \times v'$
Step 3: Solve for the unknown velocity
Simplify the equation:
$50 = -10 + 0.6\,v'$
$50 + 10 = 0.6\,v'$
$60 = 0.6\,v'$
$v' = \frac{60}{0.6} = 100\,\text{m/s}$
Hence, the other part moves upward with a velocity of $100\,\text{m/s}$.
Final Answer
The velocity of the other part is 100 m/s upwards.
