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The landing distance of each cube.
Answer to Problem 29P
The landing distance of cube is s1=34.24 cm .
The landing distance of small cube is s2=138.24 cm .
Explanation of Solution
Given:
The mass of the cube is m1 .
The mass of small cube is m2=m12 .
The height of the incline is h=30 cm .
The height of the table from the floor is H=90 cm .
Formula used:
The expression for the speed of the sliding cube is,
u1=√2gh
Here, g is the acceleration due to gravity and h is the height of sliding.
The expression for relative velocity is,
u1−u2=−(v1−v2)
Here, u1 is the initial velocity of cube, u2 is the initial velocity of the small cube, v1 is the final velocity of cube, and v2 is the final velocity of the small cube.
The expression for law of conservation of momentum is,
m1u1+m2u2=m1v1+m2v2
Here, m1 is the mass of the cube, m2 is the mass of small cube, u1 is the initial velocity of cube, u2 is the initial velocity of the small cube, v1 is the final velocity of cube, and v2 is the final velocity of the small cube.
The expression for falling time is,
t=√2Hg
Here, H is the falling height and g is the acceleration due to gravity.
The expression for landing distance is,
s=vt
Here, v is the speed and t is the time.
Calculation:
Consider the acceleration due to gravity as g=9.8 m/s2
The initial speed of the cube is,
u1=√2gh=√2×(9.8 m/s2)×(30 cm×1 m100 cm)=2.42 m/s
The initial speed of the small cube is,
u2=√2gh=√2×(9.8 m/s2)×(0)=0
The relative velocity is,
u1−u2=−(v1−v2)(2.42 m/s)−(0)=v2−v1v1=v2−2.42 …… (1)
The speed of each object using the relation of conservation of momentum is,
m1u1+m2u2=m1v1+m2v2m1(2.42 m/s)+m12(0)=m1v1+m12v22.42=v1+v222.42=(v2−2.42)+v22
4.84=3v22v2=3.23 m/s
The speed of the cube after sliding using equation (1) is,
v1=v2−2.42v1=(3.23 m/s)−2.43v1=0.8 m/s
The time of fall is,
t=√2Hgt=√2×(90 cm×1 m100 cm)9.81 m/s2t=0.428 s
The landing distance of cube is,
s1=v1ts1=(0.8 m/s)(0.428 s)s1=0.3424 m×100 cm1 ms1=34.24 cm
The landing distance of small cube is,
s2=v2ts2=(3.23 m/s)(0.428 s)s2=1.3824 m×100 cm1 ms2=138.24 cm
Conclusion:
Thus, the landing distance of cube is s1=34.24 cm and the landing distance of small cube is s2=138.24 cm .
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