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Ann [662]
3 years ago
14

A hockey player uses a 400- gram stick to apply 50 newtons of force to a 170-gram puck, causing a momentum of change 5kg x m/s.

What is the impulse experienced by the hockey puck?
Physics
1 answer:
gavmur [86]3 years ago
4 0

Answer: 5 N•s

Explanation:

An impulse will result in a change of momentum.

If the momentum changed 5 kg•m/s, you can be certain that an impulse of 5 N•s has been applied.

The units for Impulse and momentum are identical

kg•m/s = N•s

if we divide each side by s

kg•m/s/s = N•s/s

kg•m/s² = N

which is probably familiar already

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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during
Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

(900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

36300 =  (900 x v_1) + (200 x v_2)

9v_1 + 2v_2 = 363 ..............(i)

9v_1 = 363 - 2v_2

v_1=\frac{363 - 2v_2}{9}

The mathematical expression for the conservation of kinetic energy is

\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

(363-2v_2)^2+18v_2^2=195129

(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

22v_2^2-145v_2-63360=0

Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

The negative velocity is neglected.

Therefore, substituting 96 m/s for v_2 in (i), we get

v_1=\frac{363-(2 \times 96)}{9}

     = 19

Thus, only impulse of importance is used to find final velocity.

8 0
3 years ago
A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice
MrMuchimi
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

5 0
4 years ago
Read 2 more answers
Please see the picture​
Tpy6a [65]

Answer:

t.f. are you sure that's english? it looks like not a real thing

Explanation:

3 0
3 years ago
A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o
Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}.

Kinetic energy of system at first: \frac{mv^2}{2}=784\;\textbf{J}. After: \frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}. The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0
3 years ago
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