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lana66690
3 months ago
10

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

ad been placed on one of the blocks. The explosion does not harm the blocks but it doubles their total kinetic energy. After the explosion the blocks move along the x axis and the incident block?

Physics
1 answer:
Yuliya22 [3.3K]3 months ago
6 0

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

mv_o=m(v_1 + v_2)where m represents mass, v_o is the initial speed before impact, v_1 and v_2 are the velocities of the impacted object after the collision and of the originally stationary object after the impact.

5m=m(v_1 +v_2)

Thus, v_1+v_2=5

After the collision, the kinetic energy doubles, therefore:

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting the initial velocity of 5 m/s provides the equation needed to proceed.v_o

2*(5^{2})= v_1^{2} + v_2^{2}We know that v_1+v_2=5 leads to v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25. v_2=6.83 m/s

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.

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