A ball moving with velocity 2 ms^{-1} collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms^{-1}) after collision will be
0,1
1,1
1,0.5
0,2
A.
0,1
If two bodies collide head on with coefficient of restitution
A man of 50 kg mass standing in a gravity free space at height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms^{-1}. When the stone reaches the floor, the distance of the man above the floor will be
9.9 m
10.1
10 m
20 m
B.
10.1
m r = constant
The distance of the man above the floor (total height) = 10+0.1 = 10.1
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e_{1} and e_{2} respectively, the percentage error in the estimation of g is
e_{1} - e_{2}
e_{1} + 2e_{2}
e_{1}+ e_{2}
e_{1} - 2e_{2}
B.
e_{1} + 2e_{2}
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are 1 kg first part moving with a velocity of 12 ms^{-1} and 2 kg second part moving with a velocity of 8 ms^{-1}.If the third part flies off with a velocity of 4 ms^{-},its mass would be
5 kg
7 kg
17 kg
3 kg
A.
5 kg
apply the law of conservation of linear momentum.
momentum of first part = 1 x 12 = 12 kg ms^{-1}^{}
Momentum of the second part = 2 x 8 = 16 kg ms-1 '
Resultant monmentum
= [(12)^{2} +(16)^{2}]^{1/2} = 20 kg ms^{-1}
The third part should also have the same momentum.